Evaluate a limit using l'Hospital rule Evaluate $$\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}$$
I tried to apply l'Hospital rule in order to get the limit to be equal to
$$\lim_{x\to0}\frac{e^x-1}{2(e^x-\frac{x^2}{2}-x-1)^{-\frac{1}{3}}(e^x-x-1)}$$
but the new denominator has an indeterminate form itself and by repeatedly applying l'Hospital rule, it doesn't seem to help... This is where I got stuck.

Using L'Hospital's rule. Note the limit can be transformed for convenience as follows:
$$L=\frac13 \sqrt[3]{\lim_{x\to 0} \frac{(e^x-x-1)^3}{(e^x-\frac{x^2}{2}-x-1)^2}}=(L'H)=$$
$$\frac13 \sqrt[3]{\lim_{x\to 0} \frac{3(e^x-x-1)^2(e^x-1)}{2(e^x-\frac{x^2}{2}-x-1)(e^x-x-1)}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{2e^{2x}-3e^x-xe^x+1}{e^x-x-1}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{4e^{2x}-4e^x-xe^x}{e^x-1}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\sqrt[3]{\lim_{x\to 0} \frac{8e^{2x}-5e^x-xe^x}{e^x}}=\frac{1}{\sqrt[3]{18}} \cdot \sqrt[3]{3}=\frac{1}{\sqrt[3]{6}}.$$
