Help Evaluating $\lim_{x\to\infty}\left(\cos\frac{3}{x}\right)^{x^2}$ Does anyone know how to evaluate the following limit?
$$\lim_{x\to\infty}\left(\cos\frac{3}{x}\right)^{x^2}$$
I want to see a step by step solution if possible, Thank you.
 A: rewrite the limit in the form
$e^{\lim_{x \to\infty}\frac{\ln\left(\cos\left(\frac{3}{x}\right)\right)}{\frac{1}{x^2}}}$ and use the rules of L'Hospital. The searched result is $e^{-9/2}$
Sonnhard.
A: Recall that
$$
\left( 1 + \frac a r \right)^r \to e^a \text{ as }r\to\infty. 
$$
and
$$
\cos x = 1 - \frac{x^2}2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots.
$$
and in particular
$$
\cos\frac 3 x = 1 - \frac{(3/x)^2}{2} + \text{higher-degree terms}.
$$
So we have
$$
\left(1 + \dfrac{-9/2}{x^2}\right)^{x^2}\to e^{-9/2}.
$$
I posted this as a comment rather than as an answer because I didn't want to write out how to deal with the higher-degree terms.  But, goaded on by Robin Goodfellow's comment, I'm putting it here.
A: $\cos\dfrac3x=1-2\sin^2\dfrac3{2x},$
$$\lim_{x\to\infty}\left(\cos\frac3x\right)^{x^2}=\left(\lim_{x\to\infty}\left[1+\left(-2\sin^2\dfrac3{2x}\right)\right]^{--2\sin^2\dfrac3{2x}}\right)^{\left(\lim_{x\to\infty}-2x^2\sin^2\dfrac3{2x}\right)}$$
The inner limit $=e$
For the exponent set  $\dfrac3{2x}=h\iff x=\dfrac3{2h}$ 
$$\lim_{x\to\infty}-2x^2\sin^2\dfrac3{2x}=-2\cdot\dfrac94\left(\lim_{h\to0}\dfrac{\sin h}h\right)^2$$
A: $$ \lim\limits_{x\to \infty} \left(\cos\frac{3}{x}\right)^{x^2}$$
$$= \lim\limits_{x\to \infty} \exp\left(\ln\left(\cos\frac{3}{x}\right)^{x^2}\right)$$
$$= \lim\limits_{x\to \infty} \exp\left(x^2\ln\left(\cos\frac{3}{x}\right)\right)$$
Let $t=\frac{3}{x}$, then
$$\lim\limits_{t\to 0} \exp\left(\frac{9\ln\left(\cos t\right)}{t^2}\right)$$
$$=\exp\left(9\lim\limits_{t\to 0}\frac{\ln\left(\cos t\right)}{t^2}\right)$$
$$=\exp\left(9\lim\limits_{t\to 0}\frac{\cos t-1}{t^2}\right)$$
$$=\exp\left(-9\lim\limits_{t\to 0}\frac{1-\cos t}{t^2}\right)$$
$$=\exp\left(-\frac92\lim\limits_{t\to 0}\frac{t^2}{t^2}\right)$$
$$=\exp\left(-\frac92\right)$$
