Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

share|improve this question
5  
Write $x = \sqrt[5]{y}$. Look sharp. Simplify. –  Daniel Fischer Aug 19 at 23:01
1  
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 at 23:03
    
@ChrisK Thank you for your help, i like your technique :) –  egarro Aug 19 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 at 23:10
1  
@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 at 2:48

2 Answers 2

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:

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

share|improve this answer
1  
Thank you so much! –  egarro Aug 19 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 $$\lim_{x\to2}\frac{x^2-3x+2}{x^5-4x^3}=\lim_{x\to2}\frac{x^2-x-2x+2}{x^3(x^2-4)}=$$ $$=\lim_{x\to2}\frac{x(x-1)-2(x-1)}{x^3(x-2)(x+2)}=\lim_{x\to2}\frac{(x-1)(x-2)}{x^3(x-2)(x+2)}$$ $$=\lim_{x\to2}\frac{(x-1)}{x^3(x+2)}=1/32$$

share|improve this answer
    
Thanks for your help. –  egarro Aug 19 at 23:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.