Question

Evaluate: $\underset{x\to 0}{lim}\frac{e^x+\log \left(\frac{1-x}{e}\right)}{\tan x-x}$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $2$
• D. $-2$

Answer 1

Answer: (B)

$-\frac{1}{2}$

$\underset{x\to 0}{lim}\frac{e^x+\log \left(\frac{1-x}{e}\right)}{\tan x-x}$
$={lim}_{x\to 0}\frac{e^x+\log \left(1-x\right)-1}{\tan x-x}$

$=\underset{x\to 0}{lim}\frac{(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+.....)+(-x-\frac{x^2}{2}-\frac{x^3}{3}-....)-1}{(x+\frac{x^3}{3}+......)-x}$

$=\frac{\frac{1}{6}-\frac{1}{3}}{\frac{1}{3}}$
$=-\frac{1}{2}$