MCQ
The value of $\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^x} - x - 1}}{{{x^2}}}$ is
  • $0.5$
  • B
    $0$
  • C
    $1$
  • D
    $-1$

Answer

Correct option: A.
$0.5$
a
$\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}}=\frac{e^{0}-1-0}{0}=\frac{1-1}{0}=\frac{0}{0}$

This is an indeterminate type so use I'Hopital's Rule. That is, take the derivative of the top and the bottom and then find the limit of its quotient.

$\lim _{x \rightarrow 0} \frac{e^{x}-1}{2 x}=\frac{e^{0}-1}{0}=\frac{1-1}{0}=\frac{0}{0}$ This is still an indeterminate form so let's

use I'Hopital's Rule again.

$\lim _{x \rightarrow 0} \frac{e^{x}}{2}=\frac{e^{0}}{2}=\frac{1}{2}$

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