# Evaluate the following limits:

Question:

Evaluate the following limits:

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

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 $\frac{0}{0}$

[NOTE: $1-\cos x=2 \sin ^{2}(x / 2)$ ]

Formula used: $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

So, by using the above formula, we have

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

Divide numerator and denominator by $x^{2}$, we have

$\lim _{x \rightarrow 0} \frac{2 \sin ^{2}(x / 2)}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{\frac{2 \sin ^{2}\left(\frac{x}{2}\right)}{x^{2}}}{\frac{\sin ^{2} x}{x^{2}}}=\lim _{x \rightarrow 0} \frac{\frac{2 \sin ^{2}\left(\frac{x}{2}\right)}{4 \frac{x^{2}}{4}}}{\frac{\sin ^{2} x}{x^{2}}}=\frac{\frac{2}{4}}{1}=\frac{2}{4}=\frac{1}{2}$

[NOTE: $\left.\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}\right]$

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