First slide

Introduction to limits

Question

The value of $lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}}$ is equal to

Moderate

A

1/5
B

1 /6
C

1 / 4
D

1 / 2

Solution

We have,
$\begin{array}{l} lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^{4}} \\ = lim_{x \to 0} \frac{2 \sin (\frac{x + \sin x}{2}) \sin (\frac{x - \sin x}{2})}{x^{4}} \\ = 2 lim_{x \to 0} \frac{\sin (\frac{x + \sin x}{2})}{(\frac{x + \sin x}{2})} \times \frac{\sin (\frac{x - \sin x}{2})}{(\frac{x - \sin x}{2})} \times (\frac{x + \sin x}{2 x}) (\frac{x - \sin x}{2 x^{3}}) \end{array}$
$\begin{array}{l} = 2 lim_{x \to 0} \{\frac{\sin (\frac{x + \sin x}{2})}{x + \sin x} \frac{}{2}\} \times \{\frac{\sin (\frac{x - \sin x}{2})}{x - \sin x}] \end{array} \times (\frac{1}{2} + \frac{\sin x}{2 x}) (\frac{x - \sin x}{2 x^{3}})$
$\begin{array}{l} = 2 \times 1 \times 1 \times (\frac{1}{2} + \frac{1}{2}) lim_{x \to 0} \frac{x - \sin x}{2 x^{3}} \\ = 2 lim_{x \to 0} \frac{x - \sin x}{2 x^{3}} = lim_{x \to 0} \frac{x - \sin x}{x^{3}} \\ = lim_{x \to 0} \frac{x - (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \dots)}{x^{3}} = lim_{x \to 0} (\frac{1}{3} + \frac{x^{2}}{5!} \dots) = \frac{1}{3!} = \frac{1}{6} \end{array}$

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