Evaluating $\lim _{x\to \infty }\tan\left(x^2\sin\frac{\pi }{4x^2}\right)$ without L'hopital's rule

Question

How do I solve this without using L'hopital's rule?

$$\lim _{x\to \infty }\left(\tan\left(x^2\sin\left(\frac{\pi }{4x^2}\right)\right)\right)$$

Problem

EDIT:
My progress so far:

$$\lim _{x\to \infty }\left(\tan\left(x^2\sin\left(\frac{\pi }{4x^2}\right)\right)\right)$$

$$=\lim _{x\to \infty }\left(\tan\left(\frac{\sin\left(\frac{\pi }{4x^2}\right)}{\frac{1}{x^2}}\right)\right)$$

$$=\lim _{x\to \infty }\left(\tan\left(\frac{\pi}{4}*\frac{\sin\left(\frac{\pi }{4x^2}\right)}{\frac{\pi}{4x^2}}\right)\right)$$

Now I know $\frac{\sin\left(x\right)}{x} = 1$ but I don't know how to take the above expression to that form.

Answer 1

You've already taken care of one important bit, namely rewriting

$$\tan\left(x^2\sin\frac{\pi}{4x^2}\right)$$

as

$$\tan\left(\frac{\pi}{4}\cdot\frac{\sin\frac{\pi}{4x^2}}{\frac{\pi}{4x^2}}\right)$$

As you noted, $\lim_{x\to 0}\frac{\sin(x)}{x}=1$, so we should make use of this somehow. We can achieve this goal by making use of another important observation, namely that

$$\lim_{x\to\infty}\frac{\pi}{4x^2}=\frac{\pi}{4}\lim_{x\to\infty}\frac{1}{x^2}=\frac{\pi}{4}\cdot 0=0$$

Knowing this, I hope it makes sense that we should have

$$\lim_{x\to\infty}\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}=1$$

After all, as $x$ gets bigger and bigger, $\frac{\pi}{4x^2}$ gets closer and closer to $0$, and $\sin(y)/y$ gets closer and closer to $1$ as its argument gets smaller and smaller, namely $y=\frac{\pi}{4x^2}$.

Alright, so we know that

$$\lim_{x\to\infty}\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}=1$$

and thus, by the constant multiple rule for limits

$$\lim_{x\to\infty}\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)=\frac{\pi}{4}$$

But the tangent function $\tan$ is continuous near $\pi/4$, so

$$\lim_{x\to\infty}\tan\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)=\tan\left(\lim_{x\to\infty}\left(\frac{\pi}{4}\cdot\frac{\sin\left(\frac{\pi}{4x^2}\right)}{\frac{\pi}{4x^2}}\right)\right)=\tan\left(\frac{\pi}{4}\right)=1$$

We conclude that the original limit is

$$\lim_{x\to\infty}\tan\left(x^2\sin\frac{\pi}{4x^2}\right)=1$$

Answer 2

Hint:

$\lim _{x\to 0} \frac{\sin x}{x}=1$ then if $x\to\infty\ $ means $\frac{1}{x}\to0$. So basically $\lim _{x\to \infty} \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$ is also equal to $1$