Advance Calculus Limit question












3












$begingroup$


I'm trying to compute this limit without the use of L'Hopital's rule:



$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






calculus limits limits-without-lhopital







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm trying to compute this limit without the use of L'Hopital's rule:



    $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



    I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






    calculus limits limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm trying to compute this limit without the use of L'Hopital's rule:



      $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



      I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






      calculus limits limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      I'm trying to compute this limit without the use of L'Hopital's rule:



      $$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$



      I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?






      calculus limits limits-without-lhopital




      calculus limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      Foobaz John

      22.9k41552




      22.9k41552










      asked 6 hours ago









      Kevin CalderonKevin Calderon

      563




      563






















          3 Answers
          3






          active

          oldest

          votes


















          6












          $begingroup$

          Write the limit as
          $$
          lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
          $$
          and use the fact that
          $$
          lim_{xto 0+}frac{-2}{x}=-infty.
          $$
          to find that the limit equals $-1$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            A substitution can be helpful, as it transforms the expression into a rational function:




            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



              Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






              share|cite|improve this answer









              $endgroup$














                Your Answer





                StackExchange.ifUsing("editor", function () {
                return StackExchange.using("mathjaxEditing", function () {
                StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
                StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                });
                });
                }, "mathjax-editing");

                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "69"
                };
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function() {
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled) {
                StackExchange.using("snippets", function() {
                createEditor();
                });
                }
                else {
                createEditor();
                }
                });

                function createEditor() {
                StackExchange.prepareEditor({
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: true,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: 10,
                bindNavPrevention: true,
                postfix: "",
                imageUploader: {
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                },
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171288%2fadvance-calculus-limit-question%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                6












                $begingroup$

                Write the limit as
                $$
                lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                $$
                and use the fact that
                $$
                lim_{xto 0+}frac{-2}{x}=-infty.
                $$
                to find that the limit equals $-1$.






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  Write the limit as
                  $$
                  lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                  $$
                  and use the fact that
                  $$
                  lim_{xto 0+}frac{-2}{x}=-infty.
                  $$
                  to find that the limit equals $-1$.






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    Write the limit as
                    $$
                    lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                    $$
                    and use the fact that
                    $$
                    lim_{xto 0+}frac{-2}{x}=-infty.
                    $$
                    to find that the limit equals $-1$.






                    share|cite|improve this answer









                    $endgroup$



                    Write the limit as
                    $$
                    lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
                    $$
                    and use the fact that
                    $$
                    lim_{xto 0+}frac{-2}{x}=-infty.
                    $$
                    to find that the limit equals $-1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 6 hours ago









                    Foobaz JohnFoobaz John

                    22.9k41552




                    22.9k41552























                        3












                        $begingroup$

                        A substitution can be helpful, as it transforms the expression into a rational function:




                        • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                        begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                        & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                        & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                        & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                        end{eqnarray*}






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          A substitution can be helpful, as it transforms the expression into a rational function:




                          • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                          begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                          & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                          & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                          & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                          end{eqnarray*}






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            A substitution can be helpful, as it transforms the expression into a rational function:




                            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                            end{eqnarray*}






                            share|cite|improve this answer









                            $endgroup$



                            A substitution can be helpful, as it transforms the expression into a rational function:




                            • Set $y=4^{frac{1}{x}}$ and consider $y to +infty$


                            begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
                            & stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
                            & = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
                            & stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
                            end{eqnarray*}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            trancelocationtrancelocation

                            13.5k1827




                            13.5k1827























                                0












                                $begingroup$

                                $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                  Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                    Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    $$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$



                                    Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 40 mins ago









                                    Paras KhoslaParas Khosla

                                    2,758423




                                    2,758423






























                                        draft saved

                                        draft discarded




















































                                        Thanks for contributing an answer to Mathematics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid …



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171288%2fadvance-calculus-limit-question%23new-answer', 'question_page');
                                        }
                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        -

                                        Popular posts from this blog

                                        How to label and detect the document text images

                                        Image
                                        1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C
                                        Read more

                                        Vallis Paradisi

                                        Image
                                        Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para
                                        Read more

                                        Tabula Rosettana

                                        Image
                                        Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a
                                        Read more