Using sequential criterion for limits find the limits

Question

Using sequential criterion for limits show that the limits $$\lim_{x\to 0}\cos{\frac{1}{x^2}}$$ and $$\lim_{x\to \infty}x^{1+\sin{x}}$$ do not exist.

I don't know how to solve this type of problem with a suitable choice of $\{x_n\}$ and $\{y_n\}$.

Edit:

I want to consider two sequences $\{x_n\}$ and $\{y_n\}$ for each problem with an aim to see the $\lim x_n$, $\lim y_n$, $f(x_n)$ and $f(y_n)$. If both $\{x_n\}$ and $\{y_n\}$ congerge and $\lim f(x_n)\neq \lim f(y_n)$, then the limit $\lim f(x)$ does not exists. For example, what I need is here Prove that $\lim_{x \rightarrow 0} \mathrm {sgn} \sin (\frac{1}{x})$ does not exist.

or here

A suitable choice of ($\{x_n\}$ and $\{y_n\}$) is required.

Answer 1

Hints.

• For the first one, you can take

$$x_n=\frac1{\sqrt{n+1}}.$$

You have

$$\lim_{n\to\infty }x_n=0$$

but

$$\lim_{n\to\infty} \cos\left(\frac 1{{x_n}^2}\right)=\lim_{n\to\infty} \cos(n)$$

which is undefined (why?).

• For the second one, you can take

$$y_n=e^n.$$

You have

$$\lim_{n\to\infty }y_n=+\infty$$

but

$${y_n}^{1+\sin(y_n)}=e^{(1+\sin(y_n))\ln(y_n)}=e^{(1+\sin(e^n))n}.$$

And the sequence $((1+\sin(e^n))n)_n$ comes infinitely many times as close to $0$ as you want (why?), and get infinitely many times as big as you want (why?).