Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{(1-\cos 2x)(3+\cos x)}{x\tan ax}& x<0;\\ \frac{x(e^x-1)}{1-\cos x}& x>0;\end{array}$ then find the value of $a$ such that $\underset{x\to 0}{lim}f\left(x\right)$ exists

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is C $4$

$\underset{x\to 0^{-}}{lim}\frac{(1-\cos 2x)(3+\cos x)}{x\tan ax}=\underset{x\to 0^{+}}{lim}\frac{x(e^x-1)}{1-\cos x}$
$\Rightarrow \underset{x\to 0^{-}}{lim}\frac{\left(2{\sin }^{2}x\right)(3+\cos x)}{x^2\frac{\tan ax}{ax}a}=\underset{x\to 0^{+}}{lim}\frac{2x(e^x-1)}{4{\sin }^2\left(x/2\right)}$
$\Rightarrow \frac{2\times 1\times (3+1)}{a}=2\underset{x\to 0}{lim}\frac{{\left(\frac{x}{2}\right)}^2}{{\sin }^2\left(x/2\right)}\left(\frac{e^x-1}{x}\right)$
$\Rightarrow \frac{8}{a}=2\times 1$
$\Rightarrow a=4$