Advance Calculus Limit question The Next CEO of Stack OverflowLimit finding of an...

Why do we say “un seul M” and not “une seule M” even though M is a “consonne”?

Mathematica command that allows it to read my intentions

Could a dragon use its wings to swim?

What does this strange code stamp on my passport mean?

How to show a landlord what we have in savings?

MT "will strike" & LXX "will watch carefully" (Gen 3:15)?

How can the PCs determine if an item is a phylactery?

My boss doesn't want me to have a side project

Find the majority element, which appears more than half the time

Car headlights in a world without electricity

Is this a new Fibonacci Identity?

Compensation for working overtime on Saturdays

A hang glider, sudden unexpected lift to 25,000 feet altitude, what could do this?

What steps are necessary to read a Modern SSD in Medieval Europe?

How did scripture get the name bible?

logical reads on global temp table, but not on session-level temp table

Find a path from s to t using as few red nodes as possible

How to find if SQL server backup is encrypted with TDE without restoring the backup

What is the difference between 서고 and 도서관?

Upgrading From a 9 Speed Sora Derailleur?

Is it correct to say moon starry nights?

Prodigo = pro + ago?

How exploitable/balanced is this homebrew spell: Spell Permanency?

Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact



Advance Calculus Limit question



The Next CEO of Stack OverflowLimit finding of an indeterminate formI need compute a rational limit that involves rootsComplex Limit Without L'hopital'sLimit of $x^2e^x $as $x$ approaches negative infinity without using L'hopital's ruleSolving limit of radicals without L'Hopital $lim_{xto 64} dfrac{sqrt x - 8}{sqrt[3] x - 4} $Solve a limit without L'Hopital: $ lim_{xto0} frac{ln(cos5x)}{ln(cos7x)}$Limit question - L'Hopital's rule doesn't seem to workHow can I solve this limit without L'Hopital rule?Find a limit of a function W/OUT l'Hopital's rule.Compute $lim_{x rightarrow 4} frac{(2x^2 - 7x -4)}{(-x^2 + 8x - 16)}$












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


      1



      $begingroup$


      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 2 hours 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 1 hour 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

                                        Why do type traits not work with types in namespace scope?What are POD types in C++?Why can templates only be...

                                        Image
                                        What does it mean to express a gate in Dirac notation? A Strange Latex Symbol How can Republicans who favour free markets, consistently express anger when they don't like the outcome of that choice? What happened to Captain America in Endgame? How to solve constants out of the internal energy equation? How can I place the product on a social media post better? Will tsunami waves travel forever if there was no land? Unexpected email from Yorkshire Bank What is the relationship between spectral sequences and obstruction theory? How can I practically buy stocks? Will a top journal at least re
                                        Read more

                                        Will tsunami waves travel forever if there was no land?Why do tsunami waves begin with the water flowing away...

                                        Image
                                        How much cash can I safely carry into the USA and avoid civil forfeiture? Is there really no use for MD5 anymore? Is contacting this expert in the field something acceptable or would it be discourteous? How to pronounce 'C++' in Spanish How to creep the reader out with what seems like a normal person? Critique of timeline aesthetic Sci fi novel series with instant travel between planets through gates. A river runs through the gates How did Captain America manage to do this? Are Boeing 737-800’s grounded? Error message with tabularx How to make a pipeline wait for end-of-file or stop af
                                        Read more

                                        Simple Scan not detecting my scanner (Brother DCP-7055W)Brother MFC-L2700DW printer can print, can't...

                                        Image
                                        Can someone publish a story that happened to you? Will tsunami waves travel forever if there was no land? What is the strongest case that can be made in favour of the UK regaining some control over fishing policy after Brexit? Controversial area of mathematics Packing rectangles: Does rotation ever help? Is there a way to get a compiler for the original B programming language? Using a Lyapunov function to classify stability and sketching a phase portrait Do I have an "anti-research" personality? Reducing vertical space in stackrel What does KSP mean? Pulling the rope with one hand is
                                        Read more