Evaluate the following limits:

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

 

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