Embibe Experts Solutions for Exercise 7: Assignment

Let $a=\underset{x\to 0}{\lim }x\cot x$ and $b=\underset{x\to 0}{\lim }x\log x$, then 

Let $f\left(2\right)=4,f^{\text{'}}\left(2\right)=4$. Then ${Lt}_{x\to 2}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$ is

$\underset{x\to 2}{lim}\left(\frac{{\left[x\right]}^3}{3}-{\left[\frac{x}{3}\right]}^3\right)$ is (where $\left[\right]$ is G.I.F.)

If $\left\{x\right\}$ denotes the fractional part of $x$. Then $\underset{\mathrm{x}\to 0}{\lim }\frac{\left\{x\right\}}{\tan \left\{x\right\}}$ is equal to

The value of $\underset{x\to 0}{lim}\left(\frac{\sin 2x}{x}+\frac{e^x-1}{x}+\frac{\log \left(1+x\right)}{x}\right)$ is equal to

If $\alpha \mathrm{and} \beta$ be the roots of the equation $a{x}^2+bx+c=0$ then $\underset{\mathrm{x}\to \frac{1}{\mathrm{\alpha }}}{\lim }\sqrt{\frac{1-{\cos }^2\left(c{x}^2+bx+a\right)}{4{\left(1-\alpha x\right)}^2}}$ equals

If the value of $\underset{x\to 0}{\lim }\frac{{\left(1+\sin x\right)}^{\mathrm{cosec}x}-e+\left({\displaystyle \frac{\sin x}{2}}\right)e}{{\sin }^{2}x}$ is $k$, then $\left(24k-11e\right)$ is equal to

Suppose $f\left(x\right)$ is differentiable at $x=1$ and $\underset{n\to 0}{Lt}\frac{1}{n}f\left(1+n\right)=5$, then $f^{\text{'}}\left(1\right)$ equals -