Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$ The limit is:
$$
  \lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}
$$
When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:
$$
  \frac{1}{5x^{\frac45}}
$$
My question is: 


*

*How to do it without calculator.

*Show me the steps on how it's being done.

Remember that
$$
a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).
$$
Then,
$$
a-b = \frac{a^5-b^5}{a^4 + a^3b + a^2b^2 + ab^3 + b^4}
$$
Taking $a = (x+h)^{1/5}$ and $b=x^{1/5}$ we get that
\begin{align}
\lim_{h \to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} &= \lim_{h \to 0}\frac{(x+h) -x}{h(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\
&= \lim_{h \to 0}\frac{1}{(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\
&= \frac{1}{5x^{4/5}},
\end{align}
since $\lim_{h\to 0}a = \lim_{h\to 0}b = x^{1/5}$.
