Evaluate the following limits:


Evaluate the following limits:

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



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

$\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\left(\frac{1-\cos x}{\cos x}\right) \sin x}{x^{3}}=\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2} \cos x} \times \frac{\sin x}{x}$

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

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

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


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