The value of limx→0 1−cos⁡(1−cos⁡x))x4, is

A
$\frac{1}{8}$
B
$\frac{1}{2}$
C
$\frac{1}{4}$
D
none of these

Solution:

We have,

$\begin{array}{l} lim_{x \to 0} \frac{1 - \cos (1 - \cos x)}{x^{4}} = lim_{x \to 0} \frac{1 - \cos (2 \sin^{2} x / 2)}{x^{4}} \\ = lim_{x \to 0} \frac{2 \sin^{2} (\sin^{2} x / 2)}{x^{4}} = 2 lim_{x \to 0} {\{\frac{\sin (\sin^{2} x / 2)}{x^{2}}\}}^{2} \\ = 2 lim_{x \to 0} {\{\frac{\sin (\sin^{2} x / 2)}{\sin^{2} x / 2} \times \frac{\sin^{2} x / 2}{x^{2} / 4} \times \frac{1}{4}\}}^{2} = 2 {(\frac{1}{4})}^{2} = \frac{1}{8} \\ \begin{aligned} ALITER & lim_{x \to 0} \frac{1 - \cos (1 - \cos x)}{x^{4}} \\ = & lim_{x \to 0} \{\frac{1 - \cos (1 - \cos x)}{(1 - \cos x)^{2}}\} {(\frac{1 - \cos x}{x^{2}})}^{2} \\ = & (\frac{1}{2}) \times {(\frac{1}{2})}^{2} = \frac{1}{8} \end{aligned} \end{array}$

The value of $lim_{x \to 0} \frac{1 - \cos (1 - \cos x))}{x^{4}}$ , is

Solution:

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