Question

Prove that,<br /><br />$\displaystyle \lim_{x\to 0}\frac{xe^{x}-log(1+x)}{x^{2}}=\frac{3}{2}$​

Answer 1

Answer:

Given,

\lim_{x\to0}\frac{xe^x-\log(+x)}{x^2}\\\;\\=\lim_{x\to0}\frac{x(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+.......\frac{x^n}{n!})-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}......)}{x^2}\\\;\\=\lim_{x\to0}\frac{(x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+.......\frac{x^{n+1}}{n!})-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}......)}{x^2}\\\;\\=\lim_{x\to0}\frac{x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+.......\frac{x^{n+1}}{n!})+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}......}{x^2}

Answer 2

if any doubt free to ask...