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I need compute the result of this limit without l'hopital's rule, I tried different techniques but I did not get the limit, which is 1/32, I would appreciate if somebody help me. Thanks.

$$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$

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Write $x = \sqrt[5]{y}$. Look sharp. Simplify. – Daniel Fischer Aug 19 '14 at 23:01
Let $x = y^{1/5}$ and then this is equivalent to $\lim_{x \rightarrow 2} \frac{x^2 - 3x +2}{x^5 - 4x^3} = \lim \frac{2x-3}{5x^4 - 12x^2} = \frac{1}{32}$. – Chris K Aug 19 '14 at 23:03
@ChrisK Thank you for your help, i like your technique :) – egarro Aug 19 '14 at 23:06
@ChrisK How do you get that x^2-3x+2 is equal to 2x -3 and x^5 -4x^3 is 5x^4 -12x^2? – egarro Aug 19 '14 at 23:10
@egarro, they are not equal. Note that $(x^2-3x+2)' = 2x-3$ and the other case is similar. This is an application of l'Hopital's rule; I missed not using it. See the other answers for an alternative method. – Chris K Aug 20 '14 at 2:48
up vote 2 down vote accepted

$$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$

We set $y^{\frac{1}{5}}=x$

When $y \to 32, x \to 2$

So,we have:

$$\lim_{x \to 2} \frac{x^2-3x+2}{x^5-4x^3}=\lim_{x \to 2} \frac{(x-1) (x-2)}{x^3(x^2-4)}=\lim_{x \to 2} \frac{(x-1)(x-2)}{x^3(x-2)(x+2)}=\lim_{x \to 2} \frac{x-1}{x^3(x+2)}=\frac{1}{8 \cdot 4}=\frac{1}{32}$$

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Thank you so much! – egarro Aug 19 '14 at 23:11

$$\lim_{y\to32}\frac{\sqrt[5]{y^2} - 3\sqrt[5]{y} + 2}{y - 4\sqrt[5]{y^3}}$$ taking $\sqrt[5]{y}=x$ we have that $$\lim_{x\to2}\frac{x^2-3x+2}{x^5-4x^3}=\lim_{x\to2}\frac{x^2-x-2x+2}{x^3(x^2-4)}=$$ $$=\lim_{x\to2}\frac{x(x-1)-2(x-1)}{x^3(x-2)(x+2)}=\lim_{x\to2}\frac{(x-1)(x-2)}{x^3(x-2)(x+2)}$$ $$=\lim_{x\to2}\frac{(x-1)}{x^3(x+2)}=1/32$$

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Thanks for your help. – egarro Aug 19 '14 at 23:14

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