Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per hand"? I tried, but couldn't figure it out.
 A: Are you sure you got $200$ and not $100$? Is there a typo here or in your input to WolframAlpha?
We have, either applying binomial theorem, or L'Hospital, or the fundamental exponential limit $\frac{e^x-1}{x}$ that
$\frac{(1+x^5)^{10}-1}{10x^5}\to1$
$\frac{\sqrt{1+x^3}-1}{\frac{1}{2}x^3}\to1$
$\frac{\sqrt[5]{1+x^2}-1}{\frac{1}{5}x^2}\to1$
Therefore, $$\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}=\left[\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}\frac{\frac{1}{2}x^3\frac{1}{5}x^2}{10x^5}\right]\frac{10x^5}{\frac{1}{2}x^3\frac{1}{5}x^2}\to 100$$
A: \begin{align*}
&\dfrac{(1+x^{5})^{10}-1}{(\sqrt{1+x^{3}}-1)(\sqrt[5]{1+x^{2}}-1)}\\
&=\dfrac{(\sqrt{1+x^{3}}+1)((1+x^{5})-1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)}{((1+x^{3})-1)((1+x^{2})-1)}\\
&=(\sqrt{1+x^{3}}+1)((1+x^{5})^{9}+\cdots+1)((1+x^{2})^{4/5}+\cdots+1)\\
&\rightarrow 2\cdot10\cdot 5\\
&=100.
\end{align*}
A: Note that


*

*$\frac{({1+x^k})^{\alpha} - 1}{x^k} \stackrel{t=x^k}{=}\frac{(1+t)^{\alpha} - 1}{t}\stackrel{t\to 0}{\longrightarrow} \left.\left((1+t)^{\alpha}\right)' \right|_{t=0} = \alpha$
Hence,
$$\frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) } =\frac{(1+x^5)^{10} -1}{x^{5}}\cdot \frac{1}{\frac{\sqrt{1+x^3}-1}{x^3}}\cdot \frac{1}{\frac{\sqrt[5]{1+x^2}-1}{x^2}}$$
$$\stackrel{x\to 0}{\longrightarrow}10\cdot \frac{1}{\frac 12}\cdot \frac{1}{\frac 15} = 100$$
