limit and absolute absolute value problem












5












$begingroup$


$$lim_{x to -2} frac{2-|x|}{2+x}$$



If I calculate the left and right-hand limit I get different results.



Left hand side:
$$lim_{x to -2^-}frac{2+x}{2+x}=1$$



Right hand side: $$lim_{x to -2^+} frac{2-x}{2+x}=text{undefined}$$



My question is that my procedure is right or wrong?






real-analysis limits absolute-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    I get a solution from an unauthorized source what explains that it is 1
    $endgroup$
    – Jobiar Hossain
    16 hours ago










  • $begingroup$
    Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
    $endgroup$
    – Mefitico
    5 hours ago
















5












$begingroup$


$$lim_{x to -2} frac{2-|x|}{2+x}$$



If I calculate the left and right-hand limit I get different results.



Left hand side:
$$lim_{x to -2^-}frac{2+x}{2+x}=1$$



Right hand side: $$lim_{x to -2^+} frac{2-x}{2+x}=text{undefined}$$



My question is that my procedure is right or wrong?






real-analysis limits absolute-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    I get a solution from an unauthorized source what explains that it is 1
    $endgroup$
    – Jobiar Hossain
    16 hours ago










  • $begingroup$
    Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
    $endgroup$
    – Mefitico
    5 hours ago














5












5








5





$begingroup$


$$lim_{x to -2} frac{2-|x|}{2+x}$$



If I calculate the left and right-hand limit I get different results.



Left hand side:
$$lim_{x to -2^-}frac{2+x}{2+x}=1$$



Right hand side: $$lim_{x to -2^+} frac{2-x}{2+x}=text{undefined}$$



My question is that my procedure is right or wrong?






real-analysis limits absolute-value







share|cite|improve this question











$endgroup$




$$lim_{x to -2} frac{2-|x|}{2+x}$$



If I calculate the left and right-hand limit I get different results.



Left hand side:
$$lim_{x to -2^-}frac{2+x}{2+x}=1$$



Right hand side: $$lim_{x to -2^+} frac{2-x}{2+x}=text{undefined}$$



My question is that my procedure is right or wrong?






real-analysis limits absolute-value




real-analysis limits absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Mutantoe

610513




610513










asked 16 hours ago









Jobiar HossainJobiar Hossain

363




363












  • $begingroup$
    I get a solution from an unauthorized source what explains that it is 1
    $endgroup$
    – Jobiar Hossain
    16 hours ago










  • $begingroup$
    Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
    $endgroup$
    – Mefitico
    5 hours ago


















  • $begingroup$
    I get a solution from an unauthorized source what explains that it is 1
    $endgroup$
    – Jobiar Hossain
    16 hours ago










  • $begingroup$
    Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
    $endgroup$
    – Mefitico
    5 hours ago
















$begingroup$
I get a solution from an unauthorized source what explains that it is 1
$endgroup$
– Jobiar Hossain
16 hours ago




$begingroup$
I get a solution from an unauthorized source what explains that it is 1
$endgroup$
– Jobiar Hossain
16 hours ago












$begingroup$
Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
$endgroup$
– Mefitico
5 hours ago




$begingroup$
Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$.
$endgroup$
– Mefitico
5 hours ago










5 Answers
5






active

oldest

votes


















6












$begingroup$

Since $x to -2$, we can assume that $x < 0$ so that $|x| = -x$.



Then $$frac{2-|x|}{2+x} = frac{2+x}{2+x} = 1 xrightarrow{x to -2} 1$$
so the limit exists and it is equal to $1$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):



    $$
    lim_{xto-2^-}frac{2-|x|}{2+x}=lim_{xto-2^-}frac{2-|x|}{2-|x|}=lim_{xto-2^-}1=1.
    $$



    When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:



    $$
    lim_{xto-2^+}frac{2-|x|}{2+x}=lim_{xto-2^+}frac{2-|x|}{2-|x|}=lim_{xto-2^+}1=1.
    $$



    Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:



    $$lim_{xto-2}frac{2-|x|}{2+x}=1.$$



    Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=frac{2-|-2|}{2-2}=frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
      $endgroup$
      – Jobiar Hossain
      15 hours ago






    • 1




      $begingroup$
      I get it. you are amazing to explain
      $endgroup$
      – Jobiar Hossain
      15 hours ago



















    2












    $begingroup$

    Multiply and divide by $2+|x|$ and cancel:
    $$lim_{x to -2} frac{2-|x|}{2+x}=lim_{x to -2} frac{4-x^2}{(2+x)(2+|x|)}=lim_{x to -2} frac{2-x}{2+|x|}=frac{2-(-2)}{2+2}=1.$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $$lim_{xrightarrow-2}frac{2-|x|}{2+x}=lim_{xrightarrow-2}frac{2+x}{2+x}=1.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
        $endgroup$
        – Jobiar Hossain
        16 hours ago












      • $begingroup$
        @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
        $endgroup$
        – Michael Rozenberg
        16 hours ago












      • $begingroup$
        yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
        $endgroup$
        – Jobiar Hossain
        16 hours ago










      • $begingroup$
        @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
        $endgroup$
        – Michael Rozenberg
        15 hours ago












      • $begingroup$
        If you draw the graph, you will find that there is a vertical Asymptote at x=-2
        $endgroup$
        – Jobiar Hossain
        15 hours ago



















      1












      $begingroup$

      What about L'Hospital's Rule?



      Since



      $$frac{2-|-2|}{2+-2}=frac{0}{0}$$



      then



      $$frac{d}{dx}(2-|x|)=-frac{x}{|x|}, frac{d}{dx}(2+x)=1$$



      evaluate
      $frac{-frac{x}{|x|}}{1}$
      at $x=-2$
      yields



      $$frac{-frac{-2}{|-2|}}{1} = frac{-(-1)}{1}=1$$






      share|cite|improve this answer








      New contributor




      chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$













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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Since $x to -2$, we can assume that $x < 0$ so that $|x| = -x$.



        Then $$frac{2-|x|}{2+x} = frac{2+x}{2+x} = 1 xrightarrow{x to -2} 1$$
        so the limit exists and it is equal to $1$.






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          Since $x to -2$, we can assume that $x < 0$ so that $|x| = -x$.



          Then $$frac{2-|x|}{2+x} = frac{2+x}{2+x} = 1 xrightarrow{x to -2} 1$$
          so the limit exists and it is equal to $1$.






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            Since $x to -2$, we can assume that $x < 0$ so that $|x| = -x$.



            Then $$frac{2-|x|}{2+x} = frac{2+x}{2+x} = 1 xrightarrow{x to -2} 1$$
            so the limit exists and it is equal to $1$.






            share|cite|improve this answer









            $endgroup$



            Since $x to -2$, we can assume that $x < 0$ so that $|x| = -x$.



            Then $$frac{2-|x|}{2+x} = frac{2+x}{2+x} = 1 xrightarrow{x to -2} 1$$
            so the limit exists and it is equal to $1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 13 hours ago









            mechanodroidmechanodroid

            27.5k62446




            27.5k62446























                3












                $begingroup$

                When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):



                $$
                lim_{xto-2^-}frac{2-|x|}{2+x}=lim_{xto-2^-}frac{2-|x|}{2-|x|}=lim_{xto-2^-}1=1.
                $$



                When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:



                $$
                lim_{xto-2^+}frac{2-|x|}{2+x}=lim_{xto-2^+}frac{2-|x|}{2-|x|}=lim_{xto-2^+}1=1.
                $$



                Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:



                $$lim_{xto-2}frac{2-|x|}{2+x}=1.$$



                Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=frac{2-|-2|}{2-2}=frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago






                • 1




                  $begingroup$
                  I get it. you are amazing to explain
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago
















                3












                $begingroup$

                When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):



                $$
                lim_{xto-2^-}frac{2-|x|}{2+x}=lim_{xto-2^-}frac{2-|x|}{2-|x|}=lim_{xto-2^-}1=1.
                $$



                When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:



                $$
                lim_{xto-2^+}frac{2-|x|}{2+x}=lim_{xto-2^+}frac{2-|x|}{2-|x|}=lim_{xto-2^+}1=1.
                $$



                Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:



                $$lim_{xto-2}frac{2-|x|}{2+x}=1.$$



                Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=frac{2-|-2|}{2-2}=frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago






                • 1




                  $begingroup$
                  I get it. you are amazing to explain
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago














                3












                3








                3





                $begingroup$

                When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):



                $$
                lim_{xto-2^-}frac{2-|x|}{2+x}=lim_{xto-2^-}frac{2-|x|}{2-|x|}=lim_{xto-2^-}1=1.
                $$



                When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:



                $$
                lim_{xto-2^+}frac{2-|x|}{2+x}=lim_{xto-2^+}frac{2-|x|}{2-|x|}=lim_{xto-2^+}1=1.
                $$



                Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:



                $$lim_{xto-2}frac{2-|x|}{2+x}=1.$$



                Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=frac{2-|-2|}{2-2}=frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.






                share|cite|improve this answer











                $endgroup$



                When $x$ is near $-2$ approaching it from the left, $2+x$ is equivalent to $2-|x|$ (if $x<0$, then $x=-|x|$):



                $$
                lim_{xto-2^-}frac{2-|x|}{2+x}=lim_{xto-2^-}frac{2-|x|}{2-|x|}=lim_{xto-2^-}1=1.
                $$



                When $x$ is near $-2$ approaching it from the right, $2+x$ also seems to be equivalent to $2-|x|$:



                $$
                lim_{xto-2^+}frac{2-|x|}{2+x}=lim_{xto-2^+}frac{2-|x|}{2-|x|}=lim_{xto-2^+}1=1.
                $$



                Since both one-sided limits are equal to the same number, the limit exists and is equal to $1$:



                $$lim_{xto-2}frac{2-|x|}{2+x}=1.$$



                Even though the function itself is undefined at $x=-2$ because the denominator at that point is zero ($f(-2)=frac{2-|-2|}{2-2}=frac{0}{0}$), the limit of this function at $x=-2$ does exist and is equal to $1$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 5 hours ago

























                answered 16 hours ago









                Mike R.Mike R.

                2,025314




                2,025314












                • $begingroup$
                  When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago






                • 1




                  $begingroup$
                  I get it. you are amazing to explain
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago


















                • $begingroup$
                  When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago






                • 1




                  $begingroup$
                  I get it. you are amazing to explain
                  $endgroup$
                  – Jobiar Hossain
                  15 hours ago
















                $begingroup$
                When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                $endgroup$
                – Jobiar Hossain
                15 hours ago




                $begingroup$
                When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it
                $endgroup$
                – Jobiar Hossain
                15 hours ago




                1




                1




                $begingroup$
                I get it. you are amazing to explain
                $endgroup$
                – Jobiar Hossain
                15 hours ago




                $begingroup$
                I get it. you are amazing to explain
                $endgroup$
                – Jobiar Hossain
                15 hours ago











                2












                $begingroup$

                Multiply and divide by $2+|x|$ and cancel:
                $$lim_{x to -2} frac{2-|x|}{2+x}=lim_{x to -2} frac{4-x^2}{(2+x)(2+|x|)}=lim_{x to -2} frac{2-x}{2+|x|}=frac{2-(-2)}{2+2}=1.$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Multiply and divide by $2+|x|$ and cancel:
                  $$lim_{x to -2} frac{2-|x|}{2+x}=lim_{x to -2} frac{4-x^2}{(2+x)(2+|x|)}=lim_{x to -2} frac{2-x}{2+|x|}=frac{2-(-2)}{2+2}=1.$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Multiply and divide by $2+|x|$ and cancel:
                    $$lim_{x to -2} frac{2-|x|}{2+x}=lim_{x to -2} frac{4-x^2}{(2+x)(2+|x|)}=lim_{x to -2} frac{2-x}{2+|x|}=frac{2-(-2)}{2+2}=1.$$






                    share|cite|improve this answer









                    $endgroup$



                    Multiply and divide by $2+|x|$ and cancel:
                    $$lim_{x to -2} frac{2-|x|}{2+x}=lim_{x to -2} frac{4-x^2}{(2+x)(2+|x|)}=lim_{x to -2} frac{2-x}{2+|x|}=frac{2-(-2)}{2+2}=1.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 13 hours ago









                    farruhotafarruhota

                    20.2k2738




                    20.2k2738























                        1












                        $begingroup$

                        $$lim_{xrightarrow-2}frac{2-|x|}{2+x}=lim_{xrightarrow-2}frac{2+x}{2+x}=1.$$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago












                        • $begingroup$
                          @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                          $endgroup$
                          – Michael Rozenberg
                          16 hours ago












                        • $begingroup$
                          yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago










                        • $begingroup$
                          @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                          $endgroup$
                          – Michael Rozenberg
                          15 hours ago












                        • $begingroup$
                          If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                          $endgroup$
                          – Jobiar Hossain
                          15 hours ago
















                        1












                        $begingroup$

                        $$lim_{xrightarrow-2}frac{2-|x|}{2+x}=lim_{xrightarrow-2}frac{2+x}{2+x}=1.$$






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago












                        • $begingroup$
                          @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                          $endgroup$
                          – Michael Rozenberg
                          16 hours ago












                        • $begingroup$
                          yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago










                        • $begingroup$
                          @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                          $endgroup$
                          – Michael Rozenberg
                          15 hours ago












                        • $begingroup$
                          If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                          $endgroup$
                          – Jobiar Hossain
                          15 hours ago














                        1












                        1








                        1





                        $begingroup$

                        $$lim_{xrightarrow-2}frac{2-|x|}{2+x}=lim_{xrightarrow-2}frac{2+x}{2+x}=1.$$






                        share|cite|improve this answer









                        $endgroup$



                        $$lim_{xrightarrow-2}frac{2-|x|}{2+x}=lim_{xrightarrow-2}frac{2+x}{2+x}=1.$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 16 hours ago









                        Michael RozenbergMichael Rozenberg

                        103k1891195




                        103k1891195












                        • $begingroup$
                          it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago












                        • $begingroup$
                          @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                          $endgroup$
                          – Michael Rozenberg
                          16 hours ago












                        • $begingroup$
                          yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago










                        • $begingroup$
                          @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                          $endgroup$
                          – Michael Rozenberg
                          15 hours ago












                        • $begingroup$
                          If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                          $endgroup$
                          – Jobiar Hossain
                          15 hours ago


















                        • $begingroup$
                          it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago












                        • $begingroup$
                          @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                          $endgroup$
                          – Michael Rozenberg
                          16 hours ago












                        • $begingroup$
                          yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                          $endgroup$
                          – Jobiar Hossain
                          16 hours ago










                        • $begingroup$
                          @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                          $endgroup$
                          – Michael Rozenberg
                          15 hours ago












                        • $begingroup$
                          If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                          $endgroup$
                          – Jobiar Hossain
                          15 hours ago
















                        $begingroup$
                        it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                        $endgroup$
                        – Jobiar Hossain
                        16 hours ago






                        $begingroup$
                        it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0
                        $endgroup$
                        – Jobiar Hossain
                        16 hours ago














                        $begingroup$
                        @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                        $endgroup$
                        – Michael Rozenberg
                        16 hours ago






                        $begingroup$
                        @Jobiar Hossain Since $xrightarrow-2$, we can assume that $x<0$ because for $xgeq0$ it's not interesting.
                        $endgroup$
                        – Michael Rozenberg
                        16 hours ago














                        $begingroup$
                        yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                        $endgroup$
                        – Jobiar Hossain
                        16 hours ago




                        $begingroup$
                        yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit.
                        $endgroup$
                        – Jobiar Hossain
                        16 hours ago












                        $begingroup$
                        @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                        $endgroup$
                        – Michael Rozenberg
                        15 hours ago






                        $begingroup$
                        @Jobiar Hossain $xrightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $xrightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here.
                        $endgroup$
                        – Michael Rozenberg
                        15 hours ago














                        $begingroup$
                        If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                        $endgroup$
                        – Jobiar Hossain
                        15 hours ago




                        $begingroup$
                        If you draw the graph, you will find that there is a vertical Asymptote at x=-2
                        $endgroup$
                        – Jobiar Hossain
                        15 hours ago











                        1












                        $begingroup$

                        What about L'Hospital's Rule?



                        Since



                        $$frac{2-|-2|}{2+-2}=frac{0}{0}$$



                        then



                        $$frac{d}{dx}(2-|x|)=-frac{x}{|x|}, frac{d}{dx}(2+x)=1$$



                        evaluate
                        $frac{-frac{x}{|x|}}{1}$
                        at $x=-2$
                        yields



                        $$frac{-frac{-2}{|-2|}}{1} = frac{-(-1)}{1}=1$$






                        share|cite|improve this answer








                        New contributor




                        chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$


















                          1












                          $begingroup$

                          What about L'Hospital's Rule?



                          Since



                          $$frac{2-|-2|}{2+-2}=frac{0}{0}$$



                          then



                          $$frac{d}{dx}(2-|x|)=-frac{x}{|x|}, frac{d}{dx}(2+x)=1$$



                          evaluate
                          $frac{-frac{x}{|x|}}{1}$
                          at $x=-2$
                          yields



                          $$frac{-frac{-2}{|-2|}}{1} = frac{-(-1)}{1}=1$$






                          share|cite|improve this answer








                          New contributor




                          chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            What about L'Hospital's Rule?



                            Since



                            $$frac{2-|-2|}{2+-2}=frac{0}{0}$$



                            then



                            $$frac{d}{dx}(2-|x|)=-frac{x}{|x|}, frac{d}{dx}(2+x)=1$$



                            evaluate
                            $frac{-frac{x}{|x|}}{1}$
                            at $x=-2$
                            yields



                            $$frac{-frac{-2}{|-2|}}{1} = frac{-(-1)}{1}=1$$






                            share|cite|improve this answer








                            New contributor




                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



                            What about L'Hospital's Rule?



                            Since



                            $$frac{2-|-2|}{2+-2}=frac{0}{0}$$



                            then



                            $$frac{d}{dx}(2-|x|)=-frac{x}{|x|}, frac{d}{dx}(2+x)=1$$



                            evaluate
                            $frac{-frac{x}{|x|}}{1}$
                            at $x=-2$
                            yields



                            $$frac{-frac{-2}{|-2|}}{1} = frac{-(-1)}{1}=1$$







                            share|cite|improve this answer








                            New contributor




                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






                            New contributor




                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered 5 hours ago









                            chiliNUTchiliNUT

                            1113




                            1113




                            New contributor




                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.





                            New contributor





                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            chiliNUT is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






























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