Question

$\underset{x\to 0}{lim}\frac{(1+x+x^2)-e^x}{x^2}$ is equal to

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{2}$

${lim}_{x\to 0}\frac{(1+x+x^2)-e^x}{x^2}\frac{0}{0}$form

${lim}_{x\to 0}\frac{0+1+2x-e^x}{2x}=\frac{1+0-1}{0}\frac{0}{0}$form

$={lim}_{x\to 0}\frac{0+2-e^x}{2}=\frac{2-1}{2}=\frac{1}{2}$