If $f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{\tan \pi x^2+(x+1{)}^n\sin x}{x^2+(x+1{)}^n},$ then

$$\begin{gathered} \text{If}x>0,\mathrm{So}(1+x{)}^n\to \mathrm{\infty } \mathrm{as} \mathrm{n}\to \infty \\ \Rightarrow f(x)=\underset{n\to \infty }{\lim } \frac{{\displaystyle \frac{\tan \left(\pi x^2\right)}{{(x+1)}^n}}+\sin x}{\frac{x^2}{{(x+1)}^n}+1}=\frac{0+\sin x}{0+1}=\sin x \end{gathered}$$

R.H.L=0

$\text{If}x<0, \text{so }x<1 \text{so,}(1+x{)}^n\to 0 as n\to \mathrm{\infty }$

$\Rightarrow f\left(x\right)=\frac{\tan \pi x^2}{x^2}$

$L.H.L=\underset{ x\to 0-}{Lt}-\frac{\tan \left(\pi x^2\right)}{x^2}=\pi$