First slide

Introduction to limits

Question

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

Moderate

A

0
B

1
C

½
D

$\infty$

Solution

we have ,
$\begin{array}{l} lim_{x \to 0} \frac{\log \log (1 - x^{2})}{\log \log \cos x} (form \infty / \infty) \\ = lim_{x \to 0} \frac{\frac{1}{\log 1 - x^{2})} \cdot \frac{1}{1 - x^{2}} \cdot (- 2 x)}{\frac{1}{\log \cos x} \cdot \frac{1}{\cos x} \cdot (- \sin x)} \\ = 2 lim_{x \to 0} \frac{xcos xlog \cos x}{\sin x \cdot (1 - x^{2}) \log (1 - x^{2})} \\ = 2 lim_{x \to 0} \frac{x}{\sin x} \cdot lim_{x \to 0} \frac{\cos x}{1 - x^{2}} \cdot lim_{x \to 0} \frac{\log \cos x}{\log (1 - x^{2})} \\ = 2 \times 1 \times 1 \times lim_{x \to 0} \frac{\log \cos x}{\log (1 - x^{2})} (from 0 / 0) \\ = 2 lim_{x \to 0} \frac{\frac{1}{\cos x} \cdot (- \sin x)}{\frac{1}{1 - x^{2}} \cdot (- 2 x)} \\ = 2 \times \frac{1}{2} \cdot lim_{x \to 0} (\frac{\sin x}{x} \cdot \frac{1 - x^{2}}{\cos x}) = 1 \end{array}$

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