Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{x^{2+2\cos x-2}}{{}^4}& x<0\\ \frac{\sin x-\log \left(e^x\cos x\right)}{6x^2}& ,x>0\end{array}$. The value of $f\left(0\right)$, so that f is continuous is

• A. $\frac{1}{5}$
• B. $\frac{1}{3}$
• C. $\frac{1}{12}$
• D. $\frac{1}{6}$

Answer 1

Answer: (C)

$\frac{1}{12}$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\sin x-\log \left(e^x\cos x\right)}{6x^2}\frac{0}{0}$form

$=\underset{x\to 0^{+}}{lim}\frac{\cos x-\frac{{(e}^x\cos x-e^x\sin x)}{e^x\cos x\times 1}}{12x}$

$=\underset{x\to 0^{+}}{lim}\frac{\cos x-1+\tan x}{12x}$

$=\frac{1}{12}\underset{x\to 0^{+}}{lim}(x\times \frac{\cos x-1}{x^2}+\frac{\tan x}{x})$

$=\frac{1}{12}(0\times \frac{1}{12}+1)=\frac{1}{12}$