0
$\begingroup$

I am trying to solve the following limit:

\begin{equation} \lim_{x\to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} \end{equation}

I tried using $ \ln $ to get a exponential expression for the equation as follows:

\begin{equation} \text{Let} \,\, y = \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} \end{equation}

Apply $ \ln $ to bring down power and apply exponential:

\begin{equation} y = \exp\left(\frac{1}{x^{2}}\ln\left(\frac{\sin x}{x}\right)\right) \end{equation}

I have no idea how to solve the limit for $y$. From what I read, it can be done using l'Hopital rule but I am unable to get an indeterminate form no matter how I try. Could somone please assist me?

$\endgroup$

marked as duplicate by Guy Fsone, user228113, Daniel Fischer Nov 8 '17 at 13:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $$\lim_{x\to 0} (\frac{\sin x}{x})^{\frac1{x^2}} =\lim_{x\to 0}\exp\left(\frac{1}{x^2}\ln\left(\frac{\sin x -x}{x}+1\right)\right) \sim \lim_{x\to 0}\exp\left(\frac{1}{6}\frac{\ln\left(1-\frac{x^2}{6}\right)}{\frac{x^2}{6}}\right)= \exp(-\frac16)$$ Since, $$\sin x -x \sim -\frac{x^3}{6}~~~~and ~~~~ \lim_{h\to 0} \frac{\ln\left(1-h\right)}{h} = -1$$ $\endgroup$ – Guy Fsone Nov 27 '17 at 15:31
1
$\begingroup$

Marty Cohen's answer gives the most elegant solution, but since you asked for a solution using L'Hopital's rule, you should use (several applications of) it on the power:

$$\lim_{x\to0}\frac{\ln(\sin x/x)}{x^2}=\lim_{x\to0}\frac{\frac{x}{\sin x}\cdot\frac{x\cos x-\sin x}{x^2}}{2x}=\lim_{x\to0}\frac{x\cos x-\sin x}{2x^2\sin x}=\lim_{x\to0}\frac{-\sin x}{2(2\sin x+x\cos x)}$$

All of these steps are justified because we get $0/0$ each time. Iterating one last time, we get

$$\lim_{x\to0}\frac{-\cos x}{2(3\cos x-x\sin x)}$$

At last, this limit does not yield an indeterminate form. Can you take it from here?

$\endgroup$
  • $\begingroup$ I dont understand your first step. Are we allowed to apply l'hopital rule even when the expression does not exist - ln(sinx/x) is undefined as x goes to 0 $\endgroup$ – LanceHAOH Aug 16 '16 at 5:46
  • $\begingroup$ @LanceHAOH: no, the limit of $\ln(\sin x/x)$ as $x\to 0$ is zero, because the expression inside the logarithm tends to 1, and the logarithm function is continuous, and $\ln 1=0$. $\endgroup$ – symplectomorphic Aug 16 '16 at 5:48
  • $\begingroup$ Ok. sinx/x gives 0/0 when x goes to 0. Hence ln(sinx/x) will become ln(0/0). How do you differentiate something that is undefined? $\endgroup$ – LanceHAOH Aug 16 '16 at 5:49
  • $\begingroup$ @LanceHAOH: I updated my previous comment. You seem to have forgotten (or never learned) that $\sin x/x$ tends to 1 as $x$ tends to 0. You can prove this with L'Hopital's rule, too, though from some perspectives that is circular reasoning. You can also prove it by elementary geometric reasoning. $\endgroup$ – symplectomorphic Aug 16 '16 at 5:52
  • $\begingroup$ Thanks for the reminder! I understand now. $\endgroup$ – LanceHAOH Aug 16 '16 at 6:16
2
$\begingroup$

$\begin{equation} \lim_{x\to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} \end{equation} $

Taking the log, we get

$\begin{array}\\ \frac{1}{x^{2}}\ln(\frac{\sin x}{x}) &\approx \frac{1}{x^{2}}\ln(\frac{x-x^3/6+O(x^5)}{x})\\ &= \frac{1}{x^{2}}\ln(1-x^2/6+O(x^4))\\ &\approx \frac{-1}{x^{2}}(x^2/6+O(x^4))\\ &\approx \frac{-1}{6}+O(x^2)\\ \end{array} $

so the limit of the original expression is $e^{-1/6} $.

$\endgroup$
  • $\begingroup$ Please don't use \approx in the place of \sim for equivalence of functions! $\endgroup$ – Bernard Aug 16 '16 at 9:24
  • $\begingroup$ Actually, I should have used equals because of the big-oh notation. $\endgroup$ – marty cohen Aug 16 '16 at 20:55
1
$\begingroup$

Hint: Notice that, $$\lim_{x \to 0} \ln\left(\frac{\sin(x)}{x}\right) = \ln\left(\lim_{x \to 0}\frac{\sin(x)}{x}\right) = \ln\left(1\right) = 0$$

and

$$\lim_{x \to 0} x^2 = 0,$$

thus giving you an indeterminate of the form $\frac{0}{0}$.

$\endgroup$
  • $\begingroup$ But the limit does exist - see my answer. $\endgroup$ – marty cohen Aug 16 '16 at 5:08
  • 1
    $\begingroup$ I presume that LanceHOAH was in search of an indeterminate form $\frac{0}{0}$ to which to apply L'Hopital's Rule. $\endgroup$ – Yon Teh Aug 16 '16 at 5:09
  • $\begingroup$ I see what you mean, but stopping at 0/0 lead to my misinterpretation. $\endgroup$ – marty cohen Aug 16 '16 at 6:53
  • $\begingroup$ Does this provide an aswer? $\endgroup$ – Guy Fsone Nov 8 '17 at 8:28
1
$\begingroup$

Simple algebraic machinery should do.

One would suspect of course that $(1-\frac{x^2}{3!}+\frac{x^4}{5!}-...)^{\frac{1}{x^2}}\rightarrow e^{-\frac{1}{6}}$, by a simple checking of the $x^2$ coefficient.

To prove this result, observe that

$(\dfrac{1-\frac{x^2}{6}+O(x^4)}{1-\frac{x^2}{6}})^{\frac{1}{x^2}}=(1+\dfrac{O(x^4)}{1-\frac{x^2}{6}})^{\frac{1}{x^2}}$ and that $\dfrac{O(x^4)}{1-\frac{x^2}{6}}$ is itself bounded by $O(x^4)$.

Now $(1+O(x^4))^\frac{1}{x^2}\rightarrow 1$ is a well-known result.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.