# 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.

• 1) $x_n = 1/\sqrt{2n\pi}$, $y_n = 1/\sqrt{(2n+1)\pi}$; 2) $x_n = n\pi$, $y_n = (2n-1/2)\pi$. Commented Dec 4, 2019 at 13:16

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?).

• Can it be solved by considering two sequences {xn} and {yn} for each problem as I edited in the question section? Commented Feb 26, 2017 at 11:11
• @user1942348 I don't think so Commented Feb 26, 2017 at 11:19

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.