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?
Best Answer
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 $y^{5}-2^{5}=(y-2)(y^{4}+2y^{3}+2^{2}y^{2}+2^{3}y+2^{4})$