Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{e^{\frac{-x^2}{2}}-\cos x}{x\ln (1+x)\sin x(e^x-1)},& x\ne 0\\ k,& x=0\end{array}.$
If $f\left(x\right)$ is continuous at $x=0$, then $k$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{1}{8}$

Answer 1

The correct option is C $\frac{1}{12}$

$k=\underset{x\to 0}{lim}\frac{e^{\frac{-x^2}{2}}-\cos x}{x\ln (1+x)\sin x(e^x-1)}$

$k=\underset{x\to 0}{lim}\frac{e^{\frac{-x^2}{2}}-\cos x}{x^4}$

$=\underset{x\to 0}{lim}\frac{(1-\frac{x^2}{2}+\frac{x^4}{8}-...)-(1-\frac{x^2}{2}+\frac{x^4}{4!}-...)}{x^4}$

$=\frac{1}{8}-\frac{1}{24}=\frac{1}{12}$