Question

Limit x tends to 0 e^xsinx-x-x^2/ x^2+xlog(1-x)

Answer 1

Answer:

0.5

Step-by-step explanation:

$As \;\; \lim_{x \to 0} e^x=1+x\\ \\\lim_{x\to 0}sinx=1-\frac{x^2}{2}\\\\log(1-x)=-1-x-x^2\\\\Now\\\\lim_{x\to0}\frac{e^xsinx-x-\frac{x^2}{2}}{x^2+xlog(1-x)}\\\\lim_{x\to0}\;\;\frac{(1+x)(1-\frac{x^2}{2})-x-\frac{x^2}{2}}{x^2-x(1+x+x^2)}\\\\Lim_{x\to0}\frac{1-\frac{x^3}{2}+x-x^2}{x^2-x-x^2-x^3}\\\\\\=\frac{1}{2}$

Hope it Helps

Answer 2

Answer:

0.5

Step-by-step explanation:

$\begin{lgathered}As \;\; \lim_{x \to 0} e^x=1+x\\ \\\lim_{x\to 0}sinx=1-\frac{x^2}{2}\\\\log(1-x)=-1-x-x^2\\\\Now\\\\lim_{x\to0}\frac{e^xsinx-x-\frac{x^2}{2}}{x^2+xlog(1-x)}\\\\lim_{x\to0}\;\;\frac{(1+x)(1-\frac{x^2}{2})-x-\frac{x^2}{2}}{x^2-x(1+x+x^2)}\\\\Lim_{x\to0}\frac{1-\frac{x^3}{2}+x-x^2}{x^2-x-x^2-x^3}\\\\\\=\frac{1}{2}\end{lgathered}$