Originally Posted by

**Mathstud28** Who missed this one!

$\displaystyle \lim_{x\to{0}}\frac{e^x-x-\sqrt{1+x^2}}{x^3}$

Let $\displaystyle \varphi=x^3$

Then this becomes

$\displaystyle \begin{aligned}\lim_{\varphi\to{0}}\frac{e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}}{\varphi}&=\lim_{\varphi\to{0}}\frac{e^ {\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}-0}{\varphi-0}\\

&=\left(e^{\sqrt[3]{\varphi}}-\sqrt[3]{\varphi}-\sqrt{1+\sqrt[\frac{2}{3}]{\varphi}}\right)'\bigg|_{x=0}\\

&=\frac{1}{6}\end{aligned}$

You did just say no L'hopitals or Series.