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.

$$\lim_{h \to 0} \frac{(x(1+\frac{h}{x}))^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \frac{x^{1/5}(1+\frac{h}{x})^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \bigl(  x^{1/5} \cdot \frac{(1+\frac{h}{x})^{1/5} -1}{h} \bigr) = x^{1/5}\lim_{h \to 0}  \frac{(1+\frac{h}{x})^{1/5} -1}{h} =x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} $$
Since $\lim\limits_{h \to 0} \frac{1}{5} \cdot \ln{(1+\frac{h}{x})}=0$,
$$x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} = x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{h} = \\ x^{1/5}\lim_{h \to 0}  \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \lim_{h \to 0} \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{\frac{h}{x}} \cdot \lim_{h \to 0} \frac{1}{x} = x^{1/5} \cdot 1 \cdot 1/5 \cdot \frac{1}{x} = \frac{x^{1/5-1}}{5}$$
