Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{\sin ^{3} x}$

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 $: \tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\frac{\sin x-\sin x \cos x}{\cos x}=\sin x\left(\frac{1-\cos x}{\cos x}\right)$

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

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

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

Formula used: $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=1 / 2$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ or we can used $L$ hospital Rule,

So, by using the above formula, we have

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

Therefore, $\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{\sin ^{3} x}=\frac{1}{2}$

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