# Continuity and Discontinuity of $f(x)=\lim_{n\rightarrow \infty}\left(\sin x\right)^{2n}$ is

[1] If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\left(\frac{x^{2n}-1}{x^{2n}+1}\right)\;\forall x\in \mathbb{R}$. Then Continuity and Discontunity of $f(x)$ is

[2] If $f(x) = \lim_{n\rightarrow \infty}\left(\sin x\right)^{2n}\;\;\forall x\in \mathbb{R}.$ Then Continuity and Discontinuity of $f(x)$ is

$\underline{\bf{My Try::}}$ for $(1)$::

$\displaystyle f(x) = \left\{\begin{matrix} \lim_{n\rightarrow \infty} \frac{1-\frac{1}{x^{2n}}}{1+\frac{1}{x^{2n}}} = 1&, |x|>1 & \\ \lim_{n\rightarrow \infty}\frac{x^{2n}-1}{x^{2n}+1} = -1 & ,|x|<1\\ & & \end{matrix}\right.$

But I Did not Understand for $|x| = 1$, because when $|x| = 1$, Then $\displaystyle \lim_{n\rightarrow \infty}\left(\frac{x^{2n}-1}{x^{2n}+1}\right) = \frac{1^{\infty}-1}{1^{\infty}+1}$

So How Can I solve It. and in Book it is given $=0$. I Did Not Understand that

Thanks

Similarly for $(2)$, when $|\sin x| = 1$, Then $\lim_{x\rightarrow \infty}(\sin x)^{2n}=1^{\infty}$ (Which is in Indeterminant form.)

Help Required, Thanks

For your first question, make the change $y = x^2$; so you do not need to worry about the sign of $x$ since $y$ is positive. Then, if $y > 1$, your limit is $1$. If $y < 1$, your limit is $-1$.
Foir the second question, define $y =\sin^2(x)$. Since y is positive and $\sin(x) \leq 1$, then ....I let you continue hoping that this helped.
For $x \in A:=\{k\frac{\pi}{2}: k\in \mathbb{Z}\}$ we have $\sin(x)=1$ and otherwise $|\sin(x)|<1$ for $x \in \mathbb{R}\setminus A$.
Thus, we have: $$f(x) = \begin{cases} 1, & \text{if }x\in A \\ 0, & \text{if } otherwise \end{cases}$$
The reason why $f(x)=0$, in $[1]$, is because when $|x|=1$ and you consider $f_n(x)=\left(\frac{x^{2n}-1}{x^{2n}+1}\right)=0 ;\forall n \in \mathbb{N}$ that's why the limit is zero. and for $[2]$, when $\sin(x)=1$ the sequence $f_n$ equals 1 $\forall n\in \mathbb{N}$