$\underset{\mathrm{x}\to 1}{\lim }{\left(\frac{1 + \mathrm{x}}{2 + \mathrm{x}}\right)}^{\frac{\left(1 - \sqrt{\mathrm{x}}\right)}{\left(1 - \mathrm{x}\right)}}$ is equal to

• 1

• does not exist

• $\sqrt{\frac{2}{3}}$

• $\ln \left(2\right)$

The value of $\underset{\mathrm{n}\to \infty }{\lim }\left\{\frac{\sqrt{\mathrm{n} + 1} + \sqrt{\mathrm{n} + 2} + ... + \sqrt{2\mathrm{n} - 1}}{{\mathrm{n}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right\}$ is

• $\frac{2}{3}\left(2\sqrt{2} - 1\right)$

• $\frac{2}{3}\left(\sqrt{2} - 1\right)$

• $\frac{2}{3}\left(\sqrt{2} + 1\right)$

• $\frac{2}{3}\left(2\sqrt{2} + 1\right)$

Let ${\mathrm{X}}_{\mathrm{n}} = {\left(1 - \frac{1}{3}\right)}^2{\left(1 - \frac{1}{6}\right)}^2{\left(1 - \frac{1}{10}\right)}^2 ... \left(1 - \frac{1}{{\displaystyle \frac{\mathrm{n}\left(\mathrm{n} + 1\right)}{2}}}\right), \mathrm{n} \ge 2$

Then, the value of $\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{x}}^{\mathrm{n}}$ is

• 1/3

• 1/9

• 1/81

• 0

$\underset{\mathrm{n}\to \infty }{\lim }\frac{\sqrt{1} + \sqrt{2} + ... + \sqrt{\mathrm{n} - 1}}{\mathrm{n}\sqrt{\mathrm{n}}} \mathrm{is} \mathrm{equal} \mathrm{to}$

• $\frac{1}{2}$

• $\frac{1}{3}$

• $\frac{2}{3}$

• 0

Let f: $\mathrm{R} \to \mathrm{R}$ be differentiable at x = 0. If f(0) = 0 and f'(0) = 2, then the value of

$\underset{\mathrm{x}\to 0}{\lim }\frac{1}{\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right) + \mathrm{f}\left(2\mathrm{x}\right) + \mathrm{f}\left(3\mathrm{x}\right) + ... + \mathrm{f}\left(2015\mathrm{x}\right)\right]$ is

• 2015

• 0

• $2015 \times 2016$

• $2015 \times 2014$

116.

If $\underset{\mathrm{x}\to 0}{\lim }\frac{2\mathrm{asin}\left(\mathrm{x}\right) - \sin \left(2\mathrm{x}\right)}{{\tan }^3\left(\mathrm{x}\right)}$ exists and is equal to 1, then the value of $\mathrm{\alpha }$ is

• 2

• 1

• 0

• - 1

Let f(x) be a differentiable function and f'(4) = 5. Then

$\underset{\mathrm{x}\to 2}{\lim }\frac{\mathrm{f}\left(4\right) - \mathrm{f}\left({\mathrm{x}}^2\right)}{\mathrm{x} - 2}$ equals

• 0

• 5

• 20

• - 20

Let [x] denote the greatest integer less than or equal to x for any real number x. Then,

$\underset{\mathrm{n}\to \infty }{\lim }\frac{\left[\mathrm{n}\sqrt{2}\right]}{\mathrm{n}}$ is equal to

• 0

• 2

• $\sqrt{2}$

• 1

The limit of $\left[\frac{1}{{\mathrm{x}}^2} + \frac{{\displaystyle {\left(2013\right)}^{\mathrm{x}}}}{{\mathrm{e}}^{\mathrm{x}} - 1} - \frac{1}{{\mathrm{e}}^{\mathrm{x}} - 1}\right]$ as $\mathrm{x} \to 0$

• approaches $+ \infty$

• approaches $- \infty$

• is equal to ${\log }_{\mathrm{e}}\left(2013\right)$

• does not exist

The limit of $\left\{\frac{1}{\mathrm{x}}\sqrt{1 + \mathrm{x}} - \sqrt{1 + \frac{1}{{\mathrm{x}}^2}}\right\} \mathrm{as} \mathrm{x} \to 0$

• does not exist

• is equal to $\frac{1}{2}$

• is equal to 0

• is equal to 1