Question

Given, $f\left(x\right)=\{\begin{array}{cc}\frac{\tan 4x\times \cos 3x}{x}& x\ne 0\\ k& x=0\end{array}$. If $f$ is continuous at $x=0$, then $k=$ ____

• A. $0$
• B. $4$
• C. $\frac{4}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $4$

$f\left(x\right)=\left\{\begin{array}{c}\frac{\tan 4x\times \cos 3x}{x},x\ne 0\\ k,x=0\end{array}\right\}$

Given f is continuous at $x=0$

For continuity,

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\left(\frac{\tan 4x\times \cos 3x}{x}\right)=f\left(0\right)$ (as $\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(x\right)$)

$\Rightarrow \underset{x\to 0}{lim}\left(\frac{\tan \left(4x\right)}{\left(4x\right)}\right)\left(4\right).\underset{x\to 0}{lim}\cos 3x=k$ (as $\underset{x\to a}{lim}f\left(x\right).g\left(x\right)=\underset{x\to a}{lim}f\left(x\right).\underset{x\to a}{lim}g\left(x\right)$)

$\Rightarrow \left(1\right)\left(4\right).\left(1\right)=k\left(\underset{x\to 0}{lim}\frac{\tan x}{x}=1\right)$

$\Rightarrow k=4$