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.

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,
$$\frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) } =\frac{(1+x^5)^{10} -1}{x^{5}}\cdot \frac{1}{\frac{\sqrt{1+x^3}-1}{x^3}}\cdot \frac{1}{\frac{\sqrt[5]{1+x^2}-1}{x^2}}$$
$$\stackrel{x\to 0}{\longrightarrow}10\cdot \frac{1}{\frac 12}\cdot \frac{1}{\frac 15} = 100$$
