Q102 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics
$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}=$
- A$\frac{1}{2}$
- B$\frac{1}{4}$
- C$\frac{1}{6}$
- Correct answerD$\frac{1}{8}$
Solution
Given that, $\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$
$=\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \frac{(1-\cos x)^2}{x^4} \times \frac{x^4}{\sin ^4 x}$
$=\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right)^2$ $\times \lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^4$
As $x \rightarrow 0$
$(1-\cos x) \rightarrow 0$
$=\lim _{(1-\cos x) \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{x^2}$ $\times \lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^4$
$=\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times(1)^2$
$\left\{\because \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}\right.$ and $\left.\lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right\}$
$=\frac{1}{8}$
Hence, the correct answer is option: 'D'.