First slide

Introduction to limits

Question

$lim_{x \to 0} \frac{x \cos x - \log (1 + x)}{x^{2}}$ equals

Moderate

A

$1 / 2$
B

0
C

1
D

$- 1 / 2$

Solution

Using expansions of cos x and log $(1 + x)$ , the given limit is equal to
$\begin{array}{l} lim_{x \to 0} \frac{x \{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} \dots\} - \{x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} \dots\}}{x^{2}} \\ = lim_{x \to 0} (\frac{1}{2} - \frac{x}{2!} - \frac{x}{3} + \dots) = \frac{1}{2} \end{array}$

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