Calculate this limit without L'Hôpital's rule. 
Calculate $$\lim_{x\to0}\frac{(x+32)^{1/5}-2}{x}$$ without L'Hôpital's rule.

My attempt: I first rationalized the expression to get $$\left(\frac{(x+32)^{1/5}-2}{x}\right)\left(\frac{(x+32)^{1/5}+2}{(x+32)^{1/5}+2}\right)=\frac{x+28}{x((x+32)^{1/5}+2)}$$ How should I get rid of the singular $x$ in the denominator now? Should I factor something here?
 A: Let $y=(x+32)^{1/5}$. You can write the limit as $\lim_{y \to 2} \frac {y-2} {y^{5}-2^{5}}$. It is easy to write down this limit using the formula $y^{5}-2^{5}=(y-2)(y^{4}+2y^{3}+2^{2}y^{2}+2^{3}y+2^{4})$
A: Consider the function $f(x) = (x+32)^{1/5}$. By definition, $f'(0) = \lim_{x \to 0} \dfrac{(x+32)^{1/5} - 2}{x}$.
By the power and chain rules for differentiation, you can show that $f'(x) = \dfrac{1}{5}(x+32)^{-4/5}$ where $x \neq -32$.
Therefore, $\lim_{x \to 0} \dfrac{(x+32)^{1/5} - 2}{x} = f'(0) = \dfrac{1}{80}$.
A: Hint.
Making $x+32=y^5$ we have
$$
\lim_{y\to 2}\frac{y-2}{y^5-32}
$$
A: $$L=lim_{x \to 0} \dfrac{(x+32)^{\frac{1}{5}}-2}{x}$$
$$L=lim_{x \to 0} \dfrac{(x+32)^{\frac{1}{5}}-32^{\frac{1}{5}}}{(x+32) -32}$$
Therefore,
$$L=\frac{1}{5} \cdot 32^{\frac{-4}{5}}=\frac{1}{5 \cdot 2^4}$$
A: If you want detailed proof:
$\dfrac{(2^{5}(\frac{x}{32}+1))^{\frac{1}{5}}-2}{x}=\dfrac{2(\frac{x}{32}+1)^\frac{1}{5}-2}{x}$
$(1+x)^n=1+nx+\dfrac{n(n-1)}{2}\cdot x^2.....$
$\lim_{x\to0}=\dfrac{2\bigg(1+\frac{1}{5}\frac{x}{32}+\dfrac{\frac{1}{5}(\frac{1}{5}-1)}{2}\cdot\bigg(\frac{x}{32}\bigg)^2.....\bigg)-2}{x}=\lim_{x\to0}\dfrac{\frac{2x}{5\cdot32}+...}{x}=\dfrac{2}{5\cdot 32}$
A: Let $\sqrt[5]{x+32}-2=y\implies x+32=(2+y)^5$ and $x\to0\implies y\to0$  to find
$$\lim_{x\to0}\frac{(x+32)^{0.2}-2}{x}$$
$$=\lim_{y\to0}\dfrac y{(2+y)^5-32}$$
$$=\lim_{y\to0}\dfrac y{\binom51y\cdot2^4+\binom52y^2\cdot2^3+\binom53y^3\cdot2^2+\binom51y^4\cdot2+y^5}$$
$$=?$$
