Answer to Question #238304 in Calculus for moe

Question

Evaluate the limit  $\displaystyle{\lim\limits_{x \to 0} \dfrac{e^{x}-x-1}{x^2}}$ by using l'Hopital's Rule twice.

Answer 1

$\text{Evaluate the limit } \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - x - 1}}{{{x^2}}} \text{ by using l'Hopital's Rule twice} \\ \textit{Solution} \\ \text{by direct substitution, } \frac{{{e^x} - x - 1}}{{{x^2}}} = \frac{0}{0} \text{ which is an } \textit{Intermediate form}. \\ \text{Applying the l'Hopital's rule,} \implies \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - x - 1}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left( {{e^x} - x - 1} \right)}}{{\frac{d}{{dx}}\left( {{x^2}} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{2x}} \\ \text{Again, by direct substitution, } \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{2x}} = \frac{0}{0} \text{Applying the l'Hopital's rule again,} \implies \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{d}{{dx}}\left( {{e^x} - 1} \right)}}{{\frac{d}{{dx}}\left( {2x} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x}}}{2} = \frac{1}{2} \\ \text{Hence } \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - x - 1}}{{{x^2}}} = \frac{1}{2}$