Find the limit $\lim_{x \to 0}{\left(\frac{\tan x}{x}\right)}^{1/{x^2}}$ using l'Hôpital's Rule. Here is the limit I'm trying to find.
$$\lim_{x \to 0}{\left(\frac{\tan x}{x}\right)}^{1/{x^2}}$$
Now, since it takes an indeterminate form at $x=0$, I'm taking natural logarithm and trying to find the following limit.
$$
\begin{align}
L
& = \lim_{x \to 0}\ln{\bigg(\frac{\tan{x}}{x}\bigg)}^{1/{x^2}} \\
& = \lim_{x \to 0}\frac{\ln{\left(\frac{\tan{x}}{x}\right)}}{x^2}
\end{align}
$$
Now, l'Hôpital's Rule cannot be applied here since the numerator
$ f(x)
  = \ln\left({\frac{\tan{x}}{x}}\right)
$
is undefined at $x = 0$. What should be done here?
 A: For this type of problem, it is simpler to get the limit using the Maclaurin series of the tangent function and the logarithm function.  Letting $z = 1/x^2$ we can write:
$$\begin{align}
\frac{1}{x^2} \log \bigg( \frac{\tan x}{x} \bigg)
&= \frac{1}{x^2} \Bigg[ \log \tan (x) - \log(x) \Bigg] \\[6pt]
&= \frac{1}{x^2} \Bigg[ \bigg( x + \frac{x^3}{3} + \frac{2 x^5}{15} + \frac{17 x^7}{315} + \cdots \bigg) - \log(x) \Bigg] \\[6pt]
&= \frac{1}{x^2} \log \bigg( 1 + \frac{x^2}{3} + \frac{2 x^4}{15} + \frac{17 x^6}{315} + \cdots \bigg) \\[6pt]
&= z \log \bigg( 1 + \frac{1}{3z} + \frac{2}{15 z^2} + \frac{17}{315 z^3} + \cdots \bigg) \\[6pt]
&= z \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \bigg( \frac{1}{3z} + \frac{2}{15 z^2} + \frac{17}{315 z^3} + \cdots \bigg)^n  \\[6pt]
&= z \Bigg[ \frac{1}{3z} + O(z^{-2}) \Bigg] \\[6pt]
&= \frac{1}{3} + O(z^{-1}). \\[6pt]
\end{align}$$
Taking $x \rightarrow 0$ gives $z \rightarrow \infty$ which gives:
$$\frac{1}{x^2} \cdot \log \bigg( \frac{\tan x}{x} \bigg) 
= \frac{1}{3} + O(z^{-1}) \rightarrow \frac{1}{3},$$
so we have:
$$\bigg( \frac{\tan x}{x} \bigg)^{1/x^2} \rightarrow \exp(1/3) = 1.395612.$$
A: We have
$$\left(\frac{\tan x}{x}\right)^{\frac1{x^2}}= \left[\left(1+\frac{\tan x-x}{x}\right)^{\frac x{\tan x-x}}\right]^{\frac{\tan x-x}{x^3}}$$
then use $\lim_{x\to 0} (1+x)^{\frac1x}=e$ and l’Hospital for $\lim_{x\to 0} \frac{\tan x-x}{x^3}$.
A: Continuing from your last step:
\begin{align} L= \lim \limits_{x\to 0} \frac{\tan(x)-x}{x^3}\\ \end{align}
(Notice that $\lim \limits_{x\to 0} \frac{\ln(1+x)}{x}= 1$ so $\lim \limits_{x\to 0} \frac{\ln(1+(\frac{\tan(x)}{x}-1))}{\frac{\tan(x)}{x}-1}= 1$). Apply the l'Hôpital's Rule :
\begin{align} L= \lim \limits_{x\to 0} \frac{\sec{^2}(x)-1}{3x^2}= \lim \limits_{x\to 0} \frac{\tan{^2}(x)}{3x^2}= \frac{1}{3}\\ \end{align}
A: As $x \to 0 , {\tan x \over x} \to 1$ . Therefore , $\ln{\left({\tan x \over x}\right)} \to 0$ .
Therefore , ${{\ln{\left({\tan x \over x}\right)}} \over x^2 }$ is $0 \over 0$ form at $x=0$ , hence l'hopital rule can be applied .
Applying l'hopital rule in $$L= \lim_{x \to 0} {{\ln{\left({\tan x \over x}\right)}} \over x^2 } $$, we get
$$L= \lim_{x \to 0} {{{{\sec^2 x } \over \tan x } - {1 \over x }} \over 2x } = \lim_{x \to 0} {{{x {\sec^2 x }} - \tan x } \over {{2 x^2 }\tan x}}$$
Using $\lim_{x \to 0} {\tan x \over x } = 1 $ we get
$$L = \lim_{x \to 0} {{\sec^2 x -1} \over {2 x {\tan x}}}= \lim_{x \to 0} {{\tan x } \over 2x } = {1 \over 2 }$$
Therefore , $$L= {1 \over 2}$$
