Can this limit be found without using L'Hopital's Rule?

Question

$$\lim_{x\to a} \frac{\sqrt[m]{x}-\sqrt[m]{a}}{x-a}$$

Can this limit be found without using L'Hopital's rule ?

Answer 1

For natural $m\geq2$ and $a>0$ we obtain:$$\frac{\sqrt[m]x-\sqrt[m]a}{x-a}=\tfrac{\sqrt[m]x-\sqrt[m]a}{(\sqrt[m]x-\sqrt[m]a)\left(\left(\sqrt[m]x\right)^{m-1}+\left(\sqrt[m]x\right)^{m-2}\sqrt[m]a+...+\sqrt[m]x\left(\sqrt[m]a\right)^{m-2}+\left(\sqrt[m]a\right)^{m-1}\right)}\rightarrow\frac{1}{m\sqrt[m]{a^{m-1}}}$$

Answer 2

Set $y=\sqrt[m]{x}$, and $b=\sqrt[m]{a}$;

$\sqrt[m]{x}$ is a continuos function.

We can consider $\lim y \rightarrow b$,

$\lim_{y \rightarrow b}\dfrac{y-m}{y^m-b^m}$ (See Lord Shark's comment).

1) Set $y= b+h$ and let $\lim h \rightarrow 0$.

$F(h):=\dfrac{h}{(b+h)^m-b^m}$.

Binomial expansion of $(b+h)^m$, $m+1$ terms:

$(b+h)^m= b^m +mb^{m-1}h +O(h^2)$;

$\lim_{h \rightarrow 0}F(h)=\dfrac{1}{mb^{m-1}}$.