<s>[INST] Limit problem with fifth root $$\lim_{x\to 0}\frac{x^2}{\sqrt[5]{1+5x}-1-x}$$
How to solve this limit without L'Hopital rule and Taylor series?
 [/INST] We have
$$\dfrac1{a-b} = \dfrac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^5-b^5}$$
Taking $a=\sqrt[5]{1+5x}$ and $b=1+x$, we obtain that
\begin{align}
\dfrac1{\sqrt[5]{1+5x}-1-x} = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{(1+5x)-(1+x)^5}
\end{align}
Hence,
\begin{align}
\dfrac{x^2}{\sqrt[5]{1+5x}-1-x} & = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{-10-15x-5x^2-x^3}
\end{align}
Now taking the limit as $x \to 0$, we obtain the limit to be
$$L = \dfrac{1+1+1+1+1}{-10} = -\dfrac12$$
</s>