Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$

Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...

There's got to be a more simple approach.


6 Answers 6


Working with $\exp\left(\frac1{x^2}\ln\frac{\sin x}x\right)$ and l'Hopital should be fine. Since $\frac{\sin x}x\to 1$ we have a "$\frac 00$" case here:

$$\lim_{x\to 0}\frac{\ln\frac{\sin x}x}{x^2}=\lim_{x\to0}\frac{\frac x{\sin x}\frac d{dx}\frac{\sin x}x}{2x}=\lim_{x\to0}\frac{{x\cos x-\sin x}}{2x^2\sin x}$$ Apply l'Hopital again and cancel one $x$ $$\begin{align}\lim_{x\to0}\frac{{x\cos x-\sin x}}{2x^2\sin x}&=\lim_{x\to0}\frac{-x\sin x}{4x\sin x+2x^2\cos x}\\ &=\lim_{x\to0}\frac{-\sin x}{4\sin x+2x\cos x}\end{align}$$ and l'Hopital again $$\lim_{x\to0}\frac{-\sin x}{4\sin x+2x\cos x} =\lim_{x\to0}\frac{-\cos x}{4\cos x+2\cos x-2x\sin x}=-\frac16. $$ By applying the $\exp$ again, we obtain $$ \lim_{x\to 0}\left(\frac1{x^2}\ln\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-1/6}.$$


Using Taylor Series , $$\sin x=x-\frac{x^3}{3!}+O(x^5)\implies \frac{\sin x}x=1-\frac{x^2}{3!}+O(x^4)$$

$$\implies \lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=\lim_{x\to0} \left(1-\frac{x^2}{3!}+O(x^4)\right)^{\frac1{x^2}}$$


Now if we set $\displaystyle -\frac{x^2}{3!}+O(x^4)=-\frac1u,$ the inner limit reduces to $\displaystyle\lim_{u\to\infty}\left(1+\frac1u\right)^u=e$

For the exponent, $\displaystyle\lim_{x\to0}\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}=\lim_{x\to0}\left({-\frac1{3!}+O(x^2)}\right)$ as $x\ne0$ as $x\to0$


  • $\begingroup$ Can you explain the colored part please ? $$=\lim_{x\to0}\left(\left(1-\frac{x^2}{3!}+O(x^4)\right)^{\color{red}{-\frac{x^2}{3!}+O(x^4)}}\right)^{\color{red}{\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}}}$$ How does that equal 1/x^2 ? $\endgroup$
    – GinKin
    Commented Jan 9, 2014 at 14:22
  • $\begingroup$ @GinKin, please find the edited version $\endgroup$ Commented Jan 9, 2014 at 17:44
  • $\begingroup$ @BarryCipra, please have a look into the edited version $\endgroup$ Commented Jan 9, 2014 at 17:45

Here is a method beased on Taylor series. Using the Taylor series of $\frac{\sin x}{x}$, we get

$$ e^{\frac{1}{x^2}\ln(1-x^2/3!+\dots) }= e^{\frac{1}{x^2}\ln(1-t) },$$

where $t=\frac{x^2}{3!}-\frac{x^4}{5!}+\dots$. Using the Taylor series of $\ln(1-t)$, we have

$$e^{\frac{1}{x^2}\ln(1-t) }=e^{\frac{1}{x^2}(-t-t^2/2-\dots) } = e^{\frac{1}{x^2}(-(x^2/3!-x^4/5!+\dots)-(x^2/3!-x^4/5!+\dots)^2/2-\dots) }\longrightarrow_{x\to 0} e^{-\frac{1}{3!}} $$

Note: We used the following Taylor series

$$ \frac{\sin x}{x}=1-\frac{x^2}{3!}+\dots, $$

$$ \ln(1-t)=-t-\frac{t^2}{2}- \frac{t^3}{3}-\dots\,. $$

  • $\begingroup$ We've only learned Taylor. Can explain more on how did you merge two series identities ? $\endgroup$
    – GinKin
    Commented Jan 9, 2014 at 13:35
  • 1
    $\begingroup$ @GinKin: I added more material. $\endgroup$ Commented Jan 9, 2014 at 13:43
  • $\begingroup$ Thanks. There should be a plus here: $$ e^{\frac{1}{x^2}(-x^2/3!\color{red}{+}x^4/5!-\dots) }\longrightarrow_{x\to 0} e^{-\frac{1}{3!}} $$ So all the other $Xs$ in the series are nullified because of the limit. Very nice. $\endgroup$
    – GinKin
    Commented Jan 9, 2014 at 14:27
  • $\begingroup$ @GinKin: You are welcome. $\endgroup$ Commented Jan 9, 2014 at 14:30
  • $\begingroup$ Wait I think there's a problem with the signs. the $-t$ in the $\ln$ will flip all the signs of the other series so it should be: $$e^{\frac{1}{x^2}(\color{red}{+}x^2/3!\color{red}{-}x^4/5!\color{red}{+}\dots) }\longrightarrow_{x\to 0} e^{-\frac{1}{3!}}$$ $\endgroup$
    – GinKin
    Commented Jan 9, 2014 at 14:38

$$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac1{x^2}} =e^{\lim_{x\to 0}\frac{\sin x-x}{x^3}}= e^{\lim_{x\to 0}\frac{(\sin x-x)'}{(x^3)'}}=$$$$= e^{\lim_{x\to 0}\frac{\cos x-1}{3x^2}}= e^{\lim_{x\to 0}\frac{(\cos x-1)'}{(3x^2)'}}=e^{\lim_{x\to 0}\frac{-\sin x}{6x}}=e^{\frac{-1}{6}}. $$

I used a "shortcut" for $1^{\infty}$ and applied the rule of L'Hospital twice.



$\lim_{x→α}f(x)=1$ and $\lim_{x→α}g(x)= \infty $

then $$\lim_{x→α}(f(x))^{g(x)}= \lim_{x→α}(1+f(x)−1)^{g(x)}= \lim_{x→α}[[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{(f(x)-1)}]^{g(x)}= \lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$

  • $\begingroup$ Did you not forget a log on second equality? $\endgroup$
    – user88595
    Commented Jan 9, 2014 at 13:34
  • $\begingroup$ What is that shortcut ? $\endgroup$
    – GinKin
    Commented Jan 9, 2014 at 13:36
  • $\begingroup$ @GinKin. This shortcut (exposed in my answer) turn power limit of a function in the limit of a product functions. As you can see it is very useful in this situation, especially for those who are not familiar with Taylor's formula or with other non elementary-secondary school students. $\endgroup$
    – medicu
    Commented Jan 9, 2014 at 18:20
  • $\begingroup$ @GinKin. It can be easily noticed due to the elegance and aesthetics calculations using this shortcut. $\endgroup$
    – medicu
    Commented Jan 9, 2014 at 18:27
  • $\begingroup$ First line: shouldn't there be a log in the exponent ? also why is it sinx-x ? Lastly, can you explain that last move: $$\lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$ $\endgroup$
    – GinKin
    Commented Jan 11, 2014 at 12:32


$$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$

$$f(x):=\left(\frac{\sin x}x\right)^{1/{x^2}}$$

$$\ln f(x)= \dfrac{1}{x^2}\ln \left( \dfrac{\sin x}{x}-1+1\right)$$

$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\dfrac{\sin x}{x}-1}{x^2}$$

$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\sin x-x}{x^3}$$

$$\lim_{x \to 0} \ln f(x)=\lim_{x\to 0}\cancelto{1}{\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}}\times \cancelto{\dfrac{-1}{6}}{\dfrac{\sin x-x}{x^3}}=\dfrac{-1}{6}$$

$$\lim_{x \to 0} f(x)=\text{exp}(-\dfrac{1}{6})$$

Used from

Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule

Limit of a function without using L'Hôpital Rule


Easy trick

$$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac1{x^2}} =\lim_{x\to 0}\exp\left(\frac{1}{x^2}\ln\left(\frac{\sin x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{1}{6}\frac{\ln\left(1-\frac{x^2}{6}\right)}{\frac{x^2}{6}}\right)= \exp(-\frac16)$$

Given that $$\sin x -x \sim -\frac{x^3}{6}~~~~and ~~~~ \lim_{u\to 0} \frac{\ln\left(1-u\right)}{u} = -1$$


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