(\frac{\mathcal{l}n(1 + x)}{x^{2}} + \frac{x - 1}{x}) = | MATH - LIMITS | SolveeIt | Solveeit

Today, August 23

Question

$\frac{\mathcal{l}n(1 + x)}{x^{2}} + \frac{x - 1}{x}$ =

A

∞

B

$\frac{1}{2}$

C

$- \frac{1}{2}$

D

1

Answer

$\frac{1}{2}$

Explanation

Solution

$\frac { \ln ( 1 + x ) + x ^ { 2 } - x } { x ^ { 2 } }$ using expansion

ln(1 + x) = x – $\frac { x ^ { 2 } } { 2 } + \frac { x ^ { 3 } } { 3 }$ ……...