Using sequential criterion for limits to show that limits $\lim_{x\to 0}\cos{\frac{1}{x^2}}$ and $\lim_{x\to \infty}x^{1+\sin{x}}$ do not exist 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. 
 A: 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?).
A: The first; set $x_n:= \frac {1}{\sqrt {2nπ}}, y_n:=\frac {1}{\sqrt {2nπ+\frac π2}}, n\in \Bbb N$. Both sequences have limit $0$ and $\forall n\in \Bbb N: cos(\frac {1}{x_n^2})=cos2nπ=1\to 1$, $cos(\frac {1}{y_n^2})=cos(2nπ+\frac π2)=0\to 0$. Therefore the limit does not exist.
The second; get $x_n:=2nπ, y_n:=2nπ+\frac {3π}{2}, n\in\Bbb N$. Then both of them have limit $\infty$ and $x_n^{1+sinx_n}=x_nx_n^{sinx_n}=2nπ(2nπ)^{sin2nπ}=2nπ(2nπ)^0=2nπ\to 0$ and $y_n^{1+siny_n}=(2nπ+\frac {3π}{2})(2nπ+\frac {3π}{2})^{sin(2nπ+\frac {3π}{2})}=(2nπ+\frac {3π}{2})(2nπ+\frac {3π}{2})^{-1}=1\to 1$, so the limit does not exist.
