Originally Posted by

**sbhatnagar** Q2 can be easily solved by using L Hopital's rule but since calculus isnt allowed, I have used the series expansion of ln(1+x).

$\displaystyle \lim_{x \to 0}\frac{1-\frac{x}{\ln(x+1)}}{x}$

$\displaystyle =\lim_{x \to 0}\frac{\ln{(1+x)}-x}{x \ln{(x+1)}}$

$\displaystyle =\lim_{x \to 0}\frac{(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)-x}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$\displaystyle =\lim_{x \to 0}\frac{(-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}{x (x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...)}$

$\displaystyle =\lim_{x \to 0}\frac{x^2(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{x^2 (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$\displaystyle =\lim_{x \to 0}\frac{(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+...)}{ (1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...)}$

$\displaystyle =\frac{(-\frac{1}{2}+\frac{0}{3}-\frac{0}{4}+...)}{ (1-\frac{0}{2}+\frac{0}{3}-\frac{0}{4}+...)}$

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