Question

Prove that,

$\displaystyle \lim_{x\to 0}\frac{xe^{x}-log(1+x)}{x^{2}}=\frac{3}{2}$​

Answer 1

Answer:

One way to do this is to use our knowledge of some Taylor series.

$\displaystyle e^x=1+x+\frac{x^2}{2!}+\cdots\\\\\Rightarrow xe^x = x+x^2+\frac{x^3}{2!}+\cdots\\\\\text{and}\\\\\log(1+x)=x-\frac{x^2}2+\frac{x^3}{3}-\cdots$

Thus

$\displaystyle xe^x-\log(1+x)=\tfrac32x^2+\tfrac16x^3+\cdots\\\\\Rightarrow\frac{xe^x-\log(1+x)}{x^2}=\tfrac32+\tfrac16x+\cdots\\\\\Rightarrow\lim_{x\rightarrow0}\frac{xe^x-\log(1+x)}{x^2}=\frac32$

Answer 2

Answer: 3/2

Step-by-step explanation:

Using Basics:

$\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}$$=\lim_{x\to0}1+\frac{x}{2!}+\frac{x^2}{3!}+.......\frac{x^{n-1}}{n!})+\frac{1}{2}-\frac{x}{3}+\frac{x^2}{4}......\\\;\\=1+0+0+......0)+\frac{1}{2}-0+0......\\\;\\\;=1+\frac{1}{2}\\\;\\=\frac{3}{2}$$Note:-\\\;\\1)e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}............\frac{x^n}{n!}\\\;\\2)\log(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}......$

Using L' Hospital's rule,

$\lim_{x\to0}\frac{xe^x-\log(1+x)}{x^2}\\\;\\\text{Diff. denominator and Numerator with respect to x}\\\;\\=\lim_{x\to0}\frac{xe^x+e^x-\frac{1}{1+x}}{2x}\\\;\\\text{Again Diff. denominator and Numerator with respect to x}\\\;\\=\lim_{x\to0}\frac{xe^x+e^x+e^x+\frac{1}{(1+x)^2}}{2}\\\;\\\text{Putting limit}\\\;\\=\frac{0+1+1+\frac{1}{(1+0)^2}}{2}\\\;\\=\frac{1+1+1}{2}\\\;\\=\frac{3}{2}$