G Tewani Solutions for Chapter: Limits, Exercise 1: DPP

Set of all values of $x$ such that $\underset{n\to \infty }{\lim }\frac{1}{1+{\left(\frac{4{\tan }^{-1}\left(2\mathrm{\pi x}\right)}{\pi }\right)}^{4n}}$  is a non-zero and finite number, when $n\in N$ is

$\underset{x\to \infty }{\lim }\left[x-{\log }_e\left(\frac{e^x+e^{-x}}{2}\right)\right]=$
 

$\underset{x\to \infty }{\lim }\left\{{\left(e^x+{\mathrm{\pi }}^x\right)}^{\frac{1}{x}}\right\}$, (where $\{.\}$ denotes the fractional part of $x$) is equal to:
 

If $\frac{\cos x}{\sin ax}$ is a periodic function, then $\underset{m\to \infty }{\lim }\underset{n\to \infty }{\lim }\left(1+{\cos }^{2m}\left(n!\mathrm{\pi }a\right)\right)$ is equal to:
 

If $A=\underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(\sin x\right)}{{\cos }^{-1}\left(\cos x\right)}$ and $B=\underset{x\to 0}{\lim }\frac{\left[\left|x\right|\right]}{x}$, then (where $\left[.\right]$ represent greatest integer function)
 

If $f\left(x\right)=x\left(\frac{e^{\left|x\right|+\left[x\right]}-2}{\left|x\right|+\left[x\right]}\right)$, then (where $\left[·\right]$ represents the greatest integer function)

Assume that $\underset{\theta \to -1}{\lim }f\left(\theta \right)$ exists and $\frac{{\theta }^2+\theta -2}{\theta +3}\le \frac{f\left(\theta \right)}{{\theta }^2}\le \frac{{\theta }^2+2\theta -1}{\theta +3}$ holds for a certain interval containing the point, $\theta =-1$ then $\underset{\theta \to -1}{\lim }f\left(\theta \right)$:
 

Let $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{{\tan }^{-1}\left(\tan x\right)}{1+{\left({\log }_{e}x\right)}^n}, x\ne \left(2n+1\right)\frac{\mathrm{\pi }}{2}$, then