How to evaluate this given limit

Question

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?

Answer 1

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

Answer 2

\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!

Answer 3

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.