Compute $limlimits_{xto0} left(sin x + cos xright)^{1/x}$












0












$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29


















0












$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29
















0












0








0





$begingroup$


I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits







share|cite|improve this question











$endgroup$




I hit a snag while solving exponential functions whose limits are given.



Question:



$$lim_{xto0} left(sin x + cos xright)^left(1/xright)$$



My Approach:



I am using the followin relation to solve the question of these type.



$$lim_{xto0} left(1 + xright)^left(1/xright) = e qquad(2)$$



But now how should i convert my above question so that i can apply the rule as mentioned in $(2)$.



Conclusion:



First of all help will be appreciated.



Second how to solve functions of such kind in a quick method.



Thanks,



P.S.(Feel free to edit my question if you find any errors or mistakes in my question)






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 14:17









Fortox

657




657










asked Dec 31 '18 at 12:04









GingitsuneGingitsune

247




247












  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29




















  • $begingroup$
    @Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10










  • $begingroup$
    @Did , please elaborate why would the LH step be absurd?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:24






  • 2




    $begingroup$
    Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
    $endgroup$
    – Did
    Dec 31 '18 at 12:29


















$begingroup$
@Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
$endgroup$
– Did
Dec 31 '18 at 12:10




$begingroup$
@Crostul The LH step in this case would be absurd, as in every situation where one is after a limit $limlimits_{xto0}g(x)/x$ with $g$ differentiable at $0$ and $g(0)=0$.
$endgroup$
– Did
Dec 31 '18 at 12:10












$begingroup$
@Did , please elaborate why would the LH step be absurd?
$endgroup$
– Gingitsune
Dec 31 '18 at 12:24




$begingroup$
@Did , please elaborate why would the LH step be absurd?
$endgroup$
– Gingitsune
Dec 31 '18 at 12:24




2




2




$begingroup$
Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
$endgroup$
– Did
Dec 31 '18 at 12:29






$begingroup$
Because using LH to find $limlimits_{xto0} g(x)/h(x)$ when $g(0)=h(0)=0$ requires to compute $g'(0)$ and $h'(0)$. But, in the case when $h(x)=x$, one already knows that $limlimits_{xto0}g(x)/x=g'(0)$ by definition of the derivative of $g$ since $g(x)/x=(g(x)-g(0))/(x-0)$. So, invoking LH here amounts to a beautifully (?) absurd logical circle.
$endgroup$
– Did
Dec 31 '18 at 12:29












3 Answers
3






active

oldest

votes


















6












$begingroup$

In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10






  • 1




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19






  • 2




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21



















2












$begingroup$

Alternatively, take $cos x$ out:
$$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
because using the relation you want:
$$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06



















0












$begingroup$

Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



begin{eqnarray*} left(sin x + cos xright)^{1/x}
& = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
& = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
& stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
& = & e^{1+cos'(0)} = e \
end{eqnarray*}






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%2f3057643%2fcompute-lim-limits-x-to0-left-sin-x-cos-x-right1-x%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












    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21
















    6












    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21














    6












    6








    6





    $begingroup$

    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.






    share|cite|improve this answer











    $endgroup$



    In this case the best idea is take logarithm, and then use De l'Hopital. $$lim_{x to 0} frac{ln (sin x + cos x)}{x} = lim_{x to 0} frac{cos x - sin x}{sin x + cos x} = 1$$ Hence the answer is $e^{1}=e$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 12:40

























    answered Dec 31 '18 at 12:08









    CrostulCrostul

    27.8k22352




    27.8k22352












    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21


















    • $begingroup$
      See comment on main.
      $endgroup$
      – Did
      Dec 31 '18 at 12:10






    • 1




      $begingroup$
      It isn't $e^{-1}$ anyway
      $endgroup$
      – Kenny Lau
      Dec 31 '18 at 12:19






    • 2




      $begingroup$
      @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
      $endgroup$
      – Gingitsune
      Dec 31 '18 at 12:21
















    $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10




    $begingroup$
    See comment on main.
    $endgroup$
    – Did
    Dec 31 '18 at 12:10




    1




    1




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19




    $begingroup$
    It isn't $e^{-1}$ anyway
    $endgroup$
    – Kenny Lau
    Dec 31 '18 at 12:19




    2




    2




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21




    $begingroup$
    @Crostul , When apply the $lim_{xto0}$ shouldn't the answer be $1$ ?
    $endgroup$
    – Gingitsune
    Dec 31 '18 at 12:21











    2












    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06
















    2












    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06














    2












    2








    2





    $begingroup$

    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$






    share|cite|improve this answer









    $endgroup$



    Alternatively, take $cos x$ out:
    $$lim_{xto0} left(sin x + cos xright)^left(1/xright)=\
    lim_{xto0} left(cos xright)^left(1/xright)cdot lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)}cdot color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)}=color{blue}1cdot color{red}e=e,$$
    because using the relation you want:
    $$color{blue}{lim_{xto0} left(1-2sin^2 frac x2right)^left(1/xright)=\
    lim_{xto0} left[left(1+left(-2sin^2 frac x2right)right)^frac{1}{-2sin^2 frac x2}right]^{frac{-2sin^2 frac x2}{x}}=e^0=1};\
    color{red}{lim_{xto0} left(1+frac{sin x}{cos x}right)^left(1/xright)=\
    lim_{xto0} left[left(1+frac{sin x}{cos x}right)^{frac{cos x}{sin x}}right]^frac{sin x}{xcos x}=e^1=e}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 10:05









    farruhotafarruhota

    19.7k2738




    19.7k2738












    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06


















    • $begingroup$
      thanks you helped me find one more way to approach questions of this kind
      $endgroup$
      – Gingitsune
      Jan 1 at 12:06
















    $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06




    $begingroup$
    thanks you helped me find one more way to approach questions of this kind
    $endgroup$
    – Gingitsune
    Jan 1 at 12:06











    0












    $begingroup$

    Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



    begin{eqnarray*} left(sin x + cos xright)^{1/x}
    & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
    & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
    & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
    & = & e^{1+cos'(0)} = e \
    end{eqnarray*}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



      begin{eqnarray*} left(sin x + cos xright)^{1/x}
      & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
      & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
      & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
      & = & e^{1+cos'(0)} = e \
      end{eqnarray*}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



        begin{eqnarray*} left(sin x + cos xright)^{1/x}
        & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
        & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
        & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
        & = & e^{1+cos'(0)} = e \
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$



        Another standard way works with enforcing a $color{red}{1}$ in the basis and considering the resulting exponent:



        begin{eqnarray*} left(sin x + cos xright)^{1/x}
        & = & left( color{red}{1} + (color{blue}{sin x + cos x - 1})right)^{1/x}\
        & = & left (left( 1 + (color{blue}{sin x + cos x - 1})right)^{1/(color{blue}{sin x + cos x - 1})}right)^{frac{color{blue}{sin x + cos x - 1}}{x}} \
        & stackrel {x to 0}{longrightarrow} & e^{lim_{xto 0}frac{sin x + cos x - 1}{x}}\
        & = & e^{1+cos'(0)} = e \
        end{eqnarray*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 6:11









        trancelocationtrancelocation

        9,8101722




        9,8101722






























            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%2f3057643%2fcompute-lim-limits-x-to0-left-sin-x-cos-x-right1-x%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more