Question

The value of $\lim_{x \rightarrow 0}\frac{e^{x} - (1 + x)}{x^{2}}$is

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

$\lim_{x \rightarrow 0}\frac{e^{x} - (1 + x)}{x^{2}}$=$\lim_{x \rightarrow 0}\frac{(1 + x + \frac{x^{2}}{2!} + .....) - (1 + x)}{x^{2}}$

= $\lim_{x \rightarrow 0}\frac{x^{2}\left( \frac{1}{2!} + \frac{x}{3!} + \frac{x^{2}}{4!} + ..... \right)}{x^{2}}$ = $\frac{1}{2!} = \frac{1}{2}$.