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  4. If f(x) = ( x tan 2x/ sin 3x sin 5x) is continuous at x = 0, then f (0) is equal to:

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Q. If $ f(x) = \frac{ x \, tan \, 2x}{\sin \, 3x \, \sin \, 5x}$ is continuous at $x = 0$, then $f (0)$ is equal to:

Continuity and Differentiability

Solution:

Given function is :
$ f(x) = \frac{ x \, tan \, 2x}{sin \, 3x \, sin \, 5x}$
If $f(x)$ is continuous at $x = 0$, then limit
$x \rightarrow 0 $ should exist and $f (0) =\displaystyle\lim_{x \rightarrow 0} \, f(x)$
So, $f(0) = \displaystyle\lim_{x \rightarrow 0} \frac{x \, tan \, 2x}{sin \, 3x \, sin \, 5x}$
Multiplying Nr and Dr by 3,we get
$\Rightarrow \:\:\:\: f(0) = \displaystyle\lim_{x \rightarrow 0} \frac{1}{3} \left( \frac{3x}{sin \, 3x} \right) \frac{tan \, 2x}{sin \, 5x}$
$\Rightarrow \:\:\:\: f(0) = \frac{1}{3} \displaystyle\lim_{x \rightarrow 0} \left( \frac{3x}{sin \, 3x} \right) \displaystyle\lim_{x \rightarrow 0} \frac{tan \, 2x}{sin \, 5x}$
$ = \frac{1}{3} \displaystyle\lim_{x \rightarrow 0} \frac{tan \, 2x}{sin \, 5x} \:\:\:\:\:\:\: \bigg[ \because \:\:\: \displaystyle \lim_{x \rightarrow 0} \frac{x}{sin \, x} = 1 \bigg] $
$\Rightarrow \, f(0) = \frac{1}{3} \displaystyle\lim_{x \rightarrow 0} \frac{tan \, 2x}{2x} \times 2x \frac{5x}{sin \, 5x}. \frac{1}{5x} $
$ = \frac{1}{3}. \displaystyle\lim_{x \rightarrow 0} .\frac{tan \, 2x}{2x}. \displaystyle\lim_{x \rightarrow 0} . \frac{5x}{sin \, 5x}. \frac{2x}{5x} $
$ \Rightarrow \:\:\: f(0) = \frac{1}{3} .1.1. \frac{2}{5} = \frac{2}{15}$