(\displaystyle\lim\_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x... | Mathematics | SolveeIt | Solveeit

Today, October 3

Question

$\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)+1-e^x}{4x^2-9x} \ is\ equal\ to$

A

$\frac{-1}{9}$

B

$\frac{1}{9}$

C

$\frac{-1}{18}$

D

$\frac{1}{18}$

E

0

Answer

0

Explanation

Solution

The correct option is (E) : 0