How to evaluate this given limit I am trying to evaluate this limit using L' Hopital's Rule but I am getting stuck:
$\lim_{z \to \ 0} (\frac{\sin z}{z})^\frac {1}{z^2}$
My try:
Let
$w= (\frac{\sin z}{z})^\frac {1}{z^2}$
Then, $\ln w= \frac{1}{z^2}(\frac{\sin z}{z})$
$\lim_{z \to \ 0} \ln w= \lim_{z \to \ 0} \frac{1}{z^2}(\frac{\sin z}{z})$
Using the product rule for limits,
Right hand side gives a $\frac{0}{0}$ form.
After applying L Hopitals it becomes really messy due to the differentiation of  $\frac{\sin z}{z}$
term.
Is there an alternatve way to solve it? or am I making a mistake in applying L hopitals?
 A: Hint
Using  https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion,
$$\dfrac{\sin z-z}z\text{ is O}(z^2)$$
$$\left(\dfrac{\sin z}z\right)^{\dfrac1{z^2}}$$
$$=\left(1+\dfrac{\sin z-z}z\right)^{\dfrac1{z^2}}$$
$$=\left(\left(1+\dfrac{\sin z-z}z\right)^{\dfrac z{\sin z-z}}\right)^{\dfrac{\sin z -z}{z^3}}$$
Finally use $\lim_{h\to0}(1+h)^{1/h}=e$  for the exponent, use  https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion
A: \begin{gather*}
\lim _{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} =\lim _{x\rightarrow 0}\left( 1+\frac{\sin x}{x} -1\right)^{\frac{1}{x^{2}}}\\
=\lim _{x\rightarrow 0}\left( 1+\frac{\sin x-x}{x}\right)^{\frac{1}{x^{2}}}\\
As\lim _{x\rightarrow 0}\frac{\sin x-x}{x} =0\ and\ \lim _{x\rightarrow 0}\frac{1}{x^{2}} =\infty ,\\
We\ can\ use\ the\ rule\\
\lim _{x\rightarrow 0}( 1+f( x))^{g( x)} =e^{l} ,\\
where\ l=\lim _{x\rightarrow 0}( f( x) \cdotp g( x)) ,\ \\
if\ \lim _{x\rightarrow 0} f( x) =0\ and\ \lim _{x\rightarrow 0} g( x) =\infty .\\
Now,\ l=\lim _{x\rightarrow 0}\frac{\sin x-x}{x^{3}} =\lim _{x\rightarrow 0}\frac{\cos x-1}{3x^{2}} =\lim _{x\rightarrow 0}\frac{-\sin x}{6x} =\frac{-1}{6}\\
( by\ repeated\ application\ of\ LH\ rule)\\
So,\ \\
\lim _{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^{2}}} =e^{l} =e^{\frac{-1}{6}}
\end{gather*}
Hope this helps!
A: Write
$$\left(\frac {\sin x}{x}\right)^{1/x^2}=e^{\cfrac{\ln\frac{\sin x}x}{x^2}}$$
and now use L'Hospital with
$$\lim_{x\to0}\frac{\ln\frac{\sin x}x}{x^2}$$
and thereafter use continuity of the exponential function ( the limit is $\;e^{-1/6}\;$ ).
BTW:
$$\left(\frac{\ln\frac{\sin x}x}{x^2}\right)'=\frac{\frac x{\sin x}\cdot\frac{x\cos x-\sin x}{x^2}}{2x}=\frac{x\cos x-\sin x}{2x^2\sin x}$$
The next L'Hospital you use you'll get a really nice expression in the numerator...It's not that messy.
