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.

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, $$\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}=\left[\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}\frac{\frac{1}{2}x^3\frac{1}{5}x^2}{10x^5}\right]\frac{10x^5}{\frac{1}{2}x^3\frac{1}{5}x^2}\to 100$$
