G Tewani Solutions for Chapter: Limits, Exercise 2: DPP

The value of $\underset{x\to \infty }{\lim }\left(e^{\sqrt{x^4+1}}-e^{\left(x^2+1\right)}\right)$ is:
 

The value of $\underset{x\to \frac{\pi }{4}}{\lim }\frac{{\tan }^{3}x-\tan x}{\cos \left(x+\frac{\pi }{4}\right)}$ is

The value of $\underset{x\to \frac{\pi }{2}}{\lim }\frac{(1-\sin x)\left(8x^3-{\pi }^3\right)\cos x}{(\pi -2x{)}^4}$ is

The value of$\underset{x\to \infty }{\lim }\frac{{\sum }_{r=1}^{10}{\left(x+r\right)}^{2010}}{\left(x^{1006}+1\right)\left(2x^{1004}+1\right)}$ is:

If $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x^2}=a$ and $\underset{x\to 0}{\lim }\frac{f(1-\cos x)}{g\left(x\right){\sin }^{2}x}=b$, (where $b\ne 0$), then $\underset{x\to 0}{\lim }\frac{g(1-\cos 2x)}{x^4}$ is

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{x}{\sin x},& x>0\\ 2-x,& x\le 0\end{array}\right.$ and $g\left(x\right)=\left\{\begin{array}{ll}x+3,& x<1\\ x^2-2x-2,& 1\le x<2\\ x-5,& x\ge 2\end{array}\right.$. Then, the value of $\underset{x\to 0}{\lim }g\left(f\left(x\right)\right)$ is

If $k\in I$ and $n\in I$, such that $\underset{n\to \infty }{\lim }{\left(\cos \frac{k\pi }{4}\right)}^{2n}-{\left(\cos \frac{k\pi }{6}\right)}^{2n}=0$, then

If $a_n$ and $b_n$ are positive integers and $n\in \mathrm{N}$, and $a_n+\sqrt{2}{b}_n=(2+\sqrt{2}{)}^n$, then the value of $\underset{n\to \infty }{\lim }\left(\frac{a_n}{b_n}\right)$ is: