Limits and Derivatives

If N = n!(n $\left(\mathrm{n} \in \mathrm{N}, \mathrm{n} > 2\right)$, then find

$\underset{\mathrm{N}\to \infty }{\lim }\left[{\left({\log }_2\left(\mathrm{N}\right)\right)}^{- 1} + \left({\log }_3{\left(\mathrm{N}\right)}^{- 1}\right) + ... + {\log }_{\mathrm{n}}{\left(\mathrm{N}\right)}^{- 1}\right]$

Use the formula $\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{a}}^{\mathrm{x}} - 1}{\mathrm{x}} = {\log }_{\mathrm{e}}\left(\mathrm{a}\right) \mathrm{to} \mathrm{compute} \underset{\mathrm{x}\to 0}{\lim }\frac{2^{\mathrm{x}} - 1}{\sqrt{1 + \mathrm{x}}{\displaystyle }{\displaystyle -}{\displaystyle }{\displaystyle 1}}$

The value of $\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right) - 1}{{\mathrm{x}}^2}$

• 1

• $\frac{1}{2}$

• $\frac{- 1}{2}$

• 0

The value of $\underset{\mathrm{x}\to 0}{\lim }{\left(\frac{1 + 5{\mathrm{x}}^2}{1 + 3{\mathrm{x}}^2}\right)}^{\frac{1}{{\mathrm{x}}^2}}$

• e²

• e

• $\frac{1}{\mathrm{e}}$

• $\frac{1}{{\mathrm{e}}^2}$

If f(5)=7 and f'(5)=7 then $\underset{\mathrm{x}\to 5}{\lim }\frac{\mathrm{xf}\left(5\right) - 5\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x} - 5}$ is given by

• 35

• - 35

• 28

• - 28

The value of f(0) so that the function $\frac{1 - \cos \left(1 - \cos \left(\mathrm{x}\right)\right)}{{\mathrm{x}}^4}$ is continuous everywhere is

• $\frac{1}{2}$

• $\frac{1}{4}$

• $\frac{1}{6}$

• $\frac{1}{8}$

$\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left|\mathrm{x}\right|}{\mathrm{x}}$ is equal to 

• 1

• 0

• positive infinity

• does not exist

If y = ${\tan }^{-1}\sqrt{\frac{1 - \sin \left(\mathrm{x}\right)}{1 + \sin \left(\mathrm{x}\right)}}$, then the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at x = $\frac{\mathrm{\pi }}{6}$ is

• - $\frac{1}{2}$

• $\frac{1}{2}$

• 1

• - 1

The value of the limit $\underset{\mathrm{x}\to 1}{\lim }\frac{\sin \left({\mathrm{e}}^{\mathrm{x} - 1} - 1\right)}{\log \left(\mathrm{x}\right)}$

• 0

• e

• $\frac{1}{\mathrm{e}}$

• 1

Let f(x) = $\frac{\sqrt{\mathrm{x} + 3}}{\mathrm{x} + 1}$, then the value of $\underset{\mathrm{x}\to - 3 - 0}{\lim }\mathrm{f}\left(\mathrm{x}\right)$ is

• 0

• does not exist

• $\frac{1}{2}$

• $- \frac{1}{2}$