Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^{2} 2 x}$
Solution:
To Find: Limits
NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.
In this Case, indeterminate Form is $\frac{0}{0}$
Formula used: $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Divide numerator and denominator by $x^{2}$, we have
So, by using the above formula, we have
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^{2} 2 x}=\lim _{x \rightarrow 0} \frac{\frac{[1-\cos x]}{x^{2}}}{\frac{\sin ^{2} 2 x}{x^{2}}}=\frac{1}{2}$
Therefore, $\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^{2} 2 x}=\frac{1}{2}$