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.
 A: Recall how the derivative of a function $f(x)$ is defined:
$$
f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.\tag{1}
$$
In the context of your problem, it is fairly clear that we are dealing with $(1)$ where $f(x)=x^{1/5}$, giving us
$$
f'(x) = \lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}.\tag{2}
$$
I assume you have encountered the Power Rule in calculus. Using this, we can see how to calculate $(2)$:
$$
\lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} = \frac{d}{dx}x^{1/5} = \frac{1}{5}x^{-4/5}=\frac{1}{5x^{4/5}},
$$
as desired.
A: The expression you have given is the definition for derivative of $x^{1/5}$ with respect to $x$.
$$\lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}=\frac{d}{dx}x^{1/5}=\frac{1}{5}x^{1/5-1}=\frac{1}{5}x^{-4/5}=\frac{1}{5x^{4/5}}$$
A: I guess you don't know (yet) about derivatives. Of course the answer to your question depends a bit on what you know already. Here is an answer relying on the generalized binomial theorem
$$ (a+b)^y = \sum_{i=0}^\infty \binom{y}{i} a^i b^{y-i} \tag{1}$$
with $$\binom{y}{i}= \frac{y(y-1)\cdots(y-i+1)}{i!}. $$
In your case, you have $a=h$, $b=x$, and $y=1/5$ which yields 
$$(h+x)^{1/5}-x^{1/5} = \sum_{i=1}^\infty \binom{1/5}{i} h^i x^{1/5-i}.$$
As a result, you can write
$$ \frac{(h+x)^{1/5}-x^{1/5}}{h} = \sum_{i=1}^\infty \binom{1/5}{i} h^{i-1} x^{1/5-i}.$$
Now under the limit $h\to 0$ only the term with $i=1$ survives and you simply have to evaluate
$$\lim_{h\to0}\frac{(h+x)^{1/5}-x^{1/5}}{h} = \binom{1/5}{1} x^{1/5-1}
= \frac15 x^{-4/5},$$
A: use the binomial theorem in the form $$(BIG + small)^n = BIG^n + nBIG^{n-1}small+ \cdots$$
here is how it is applied: $(x + h)^{1/5} = x^{1/5} + \frac{1}{5}x^{4/5}h + \cdots$
so that $\dfrac{(x+h)^{1/5} - x^{1/5}}{h} = \frac{1}{5}x^{4/5}+ \cdots$ and in the limit you $\frac{1}{5}x^{4/5}$
A: $$\lim_{h \to 0} \frac{(x(1+\frac{h}{x}))^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \frac{x^{1/5}(1+\frac{h}{x})^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \bigl(  x^{1/5} \cdot \frac{(1+\frac{h}{x})^{1/5} -1}{h} \bigr) = x^{1/5}\lim_{h \to 0}  \frac{(1+\frac{h}{x})^{1/5} -1}{h} =x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} $$
Since $\lim\limits_{h \to 0} \frac{1}{5} \cdot \ln{(1+\frac{h}{x})}=0$,
$$x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} = x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{h} = \\ x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \lim_{h \to 0} \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{\frac{h}{x}} \cdot \lim_{h \to 0} \frac{1}{x} = x^{1/5} \cdot 1 \cdot 1/5 \cdot \frac{1}{x} = \frac{x^{1/5-1}}{5}$$
A: 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 $\lim_{h\to 0}a = \lim_{h\to 0}b = x^{1/5}$.
