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.
 A: Use L'Hospital's Rule once to show that $$\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}=\frac{1}{2}\tag{1}$$ and using this limit and L'Hospital's Rule once more show that $$\lim_{x\to 0}\dfrac{e^{x}-1-x-\dfrac{x^{2}}{2}}{x^{3}}=\frac{1}{6}\tag{2}$$ Now divide the numerator and denominator of the original expression (whose limit is to be evaluated here) by $x^{2}$ (for denominator write $x^{2}=(x^{3})^{2/3}$) and take limits (and use limits $(1),(2)$) to get the answer as $$\dfrac{\dfrac{1}{2}}{3\left(\dfrac{1}{6}\right) ^{2/3}}=\frac{1}{\sqrt[3]{6}}$$ One must always use certain algebraic manipulation before the application of advanced techniques like L'Hospital's Rule or Taylor series unless the question is especially suited to these techniques. Your direct application of L'Hospital's Rule only leads to complicated expressions. 
A: You can use $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)$
$$\qquad{\lim_{x\to0} \frac{e^x-x-1}{3(e^x-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-x-1}{3((1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))-\frac{x^2}{2}-x-1)^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+o(x^5))}{3((\frac{x^3}{6}+\frac{x^4}{24}+o(x^5)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3(x^3(\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{x^2(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3x^{3\times\frac{2}{3}}((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
\lim_{x\to0} \frac{(\frac{1}{2}+\frac{x}{6}+\frac{x^2}{24}+o(x^3))}{3((\frac{1}{6}+\frac{x}{24}+o(x^2)))^{\frac{2}{3}}}=\\
 \frac{\frac{1}{2}}{3(\frac{1}{6})^{\frac{2}{3}}}}
$$
A: Using L'Hospital's rule. Note the limit can be transformed for convenience as follows:
$$L=\frac13 \sqrt[3]{\lim_{x\to 0} \frac{(e^x-x-1)^3}{(e^x-\frac{x^2}{2}-x-1)^2}}=(L'H)=$$
$$\frac13 \sqrt[3]{\lim_{x\to 0} \frac{3(e^x-x-1)^2(e^x-1)}{2(e^x-\frac{x^2}{2}-x-1)(e^x-x-1)}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{2e^{2x}-3e^x-xe^x+1}{e^x-x-1}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\cdot \sqrt[3]{\lim_{x\to 0} \frac{4e^{2x}-4e^x-xe^x}{e^x-1}}=(L'H)=$$
$$\frac{1}{\sqrt[3]{18}}\sqrt[3]{\lim_{x\to 0} \frac{8e^{2x}-5e^x-xe^x}{e^x}}=\frac{1}{\sqrt[3]{18}} \cdot \sqrt[3]{3}=\frac{1}{\sqrt[3]{6}}.$$
