$\underset{x→0}{\lim}\begin{Bmatrix}\frac{\log_e(1+x)}{x^2}+\frac{x-1}{x}\end{Bmatrix}=$
Answer & explanation
Correct answer: option 2
Required limit = $\underset{x→0}{\lim}\frac{\log_e(1+x)+x^2-x}{x^2}=\underset{x→0}{\lim}\frac{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-.....\right)+x^2-x}{x^2}$
$=\underset{x→0}{\lim}\frac{\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....}{x^2}=\underset{x→0}{\lim}\frac{x^2\left(\frac{1}{2}+\frac{x}{3}-.....\right)}{x^2}=\frac{1}{2}$