_ x 0 1- x ^2 x = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^2 x}=$

✓
$\frac{1}{2}$
B
$-\frac{1}{2}$
C
2
D
-2

✓

Answer

Correct option: A.

$\frac{1}{2}$

(A)
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^2 x}=\frac{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}}{\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2}}=\frac{\frac{1}{2}}{1^2}=\frac{1}{2}$
Alternate method:
Annly L-Hosnital's rule two times

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