Calculate limit with a lot of roots $$ \lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[\Large3]{1+5x}}{\sqrt[\Large5]{1+x}-\sqrt[\Large5]{1+2x}}$$
Multiplication by conjugate hasn't worked. I need a hint to start.
 A: Applying L'Hospital's rule, we get
$$\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[3]{1+5x}}{\sqrt[5]{1+x}-\sqrt[5]{1+2x}}$$
$$=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{1+2x}}-\frac{5}{3\sqrt[2/3]{1+5x}}}{\frac{1}{5\sqrt[4/5]{1+x}}-\frac{2}{5\sqrt[4/5]{1+2x}}}$$
$$=\frac{\frac{2}{2}-\frac{5}{3}}{\frac{1}{5}-\frac{2}{5}}=\frac{\frac{-2}{3}}{\frac{-1}{5}}=\color{red}{\frac{10}{3}}$$
A: Using Taylor series: when $x\to0$, we have, for any fixed  $\alpha$, $(1+x)^\alpha = 1+\alpha x + o(x)$.
In your case,
$$
\frac{(1+2x)^{1/2} - (1+5x)^{1/3}}{(1+x)^{1/5}-(1+2x)^{1/5}} =
\frac{1+x + o(x) - (1+\frac{5}{3}x + o(x)))}{1+\frac{1}{5}x + o(x) -(1+\frac{2}{5}x + o(x))} = (\cdots)
$$
Edit: in a subsequent comment, it turns out the OP is not allowed to use Taylor series. I'm leaving this here, however, as it may be useful to other readers who don't have this constraint. (?)
A: Just using the definition of the derivative: Let $a(x) = (1+2x)^{1/2},$ $ b(x) = (1+5x)^{1/3},$ $ c(x) = (1+x)^{1/5} ,d(x) = (1+2x)^{1/5}.$ The expression equals
$$\frac{a(x) - b(x)}{c(x) - d(x)} = \frac{(a(x)-a(1)) - (b(x)-b(1))}{(c(x)-c(1)) - (d(x)-d(1))}$$ $$ = \frac{(a(x)-a(1))/(x-1) - (b(x)-b(1))/(x-1)}{(c(x)-c(1))/(x-1) - (d(x)-d(1))/(x-1)}.$$
By definition of the derivative, the above has limit
$$\frac{a'(1)-b'(1)}{c'(1)-d'(1)},$$
an easy computation.
