| Date | May 2007 | Marks available | 3 | Reference code | 07M.1.hl.TZ0.2 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Calculate | Question number | 2 | Adapted from | N/A |
Question
Calculate the following limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}\) .
Calculate the following limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} - 1}}{{\ln (1 + x) - x}}\) .
Markscheme
\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{1}\) 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]
