Question

$\underset{x\to 0}{lim}\frac{sinx+log(1-x)}{x^2}$

• A. 0
• B.
• C.
• D. e

Answer 1

The correct option is C

We have, $\underset{x\to 0}{lim}\frac{sinx+log(1-x)}{x^2}$

At x=0 the value of the function is $\frac{0}{0}$ form.

We know, the expression of $sinx\&log(1-x)$

$\underset{x\to 0}{lim}\frac{(x-\frac{x^3}{3!}+\frac{x^5}{5!}.......)+(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}.......)}{x^2}$

x and -x will cancel out each other

$\underset{x\to 0}{lim}\frac{(\frac{-x^3}{3!}+\frac{x^5}{5!}.......)+(-\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}.......)}{x^2}$

$\underset{x\to 0}{lim}(\frac{-x}{3!}+\frac{x^3}{5!}.......)+(-\frac{1}{2}-\frac{x}{3}+\frac{x^2}{4}.......)$

$=0-\frac{1}{2}-0$

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

Option C is correct