Question

The value of $\underset{x\to 0}{lim}\frac{e^x+\log (1+x)-(1-x{)}^{-2}}{x^2}$ equal to

• A. $0$
• B. $-3$
• C. $-1$
• D. $\infty$

Answer 1

Answer: (B)

$-3$

Let $L=\underset{x\to 0}{lim}\frac{e^x+\log (1+x)-(1-x{)}^{-2}}{x^2}$ .......... [It reduces to $\frac{0}{0}$ form by substituting the limit]

By L' Hospital rule, we get

$L=\underset{x\to 0}{lim}\frac{e^x+(1+x{)}^{-1}-2(1-x{)}^{-3}}{2x}$ ..... $\left[\frac{0}{0}\text{form}\right]$

Again By L' Hospital's rule, we get

$L=\underset{x\to 0}{lim}\frac{e^x-(1+x{)}^{-2}-6(1-x{)}^{-4}}{2}$

$=\frac{e^0-1-6}{2}=-3$