Question

The value of $\underset{x\to 0}{lim}\frac{e(1+x{)}^2+(e+x\left)\ell n\right(e+x)-2e}{(1-\cos x+\sin x)}$

• A. $2e^2$
• B. $2(e+1)$
• C. $4e^2$
• D. 4

Answer 1

The correct option is B $2(e+1)$

Given limit is of $\frac{0}{0}$ form. So using L'Hopital's rule:

$\underset{x\to 0}{lim}\frac{e(x+1{)}^2+(e+x\left)\ell n\right(e+x)-2e}{(1-\cos x+\sin x)}$

$=\underset{x\to 0}{lim}\frac{2e(x+1)+(e+x)/(e+x)+ln(e+x)}{(\sin x+\cos x)}=\frac{2e(0+1)+e/e+lne}{0+1}=\frac{2e+2}{1}=2(1+e)$