L'Hopital rule to solve $ \lim\limits_{x\to 0} \left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} $

Question

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?

Answer 1

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?

Answer 2

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 $\frac{0}{0}$.

Answer 3

$\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} $.

Answer 4

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.