Calculate the following limit

$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}$ .

Calculate the following limit

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} - 1}}{{\ln (1 + x) - x}}$ .

$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{1}$     M1A1

$= \ln 2$     A1

[3 marks]

EITHER

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} - 1}}{{\ln (1 + x) - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}}}{{\frac{1}{{1 + x}} - 1}}$     M1A1

$= \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}(1 + x)}}{{ - x}}$     A1A1

$= - 3$     A1

OR

${(1 + {x^2})^{\frac{3}{2}}} - 1 = 1 + \frac{3}{2}{x^2} +  \ldots  - 1 = \frac{3}{2}{x^2} +  \ldots$     M1A1

$\ln (1 + x) - x = x - \frac{1}{2}{x^2} +  \ldots  - x = - \frac{1}{2}{x^2} +  \ldots$     M1A1

Limit $= - 3$     A1

[5 marks]