# Another limits by change of variable problem

• Jul 12th 2007, 11:42 AM
starswept
Another limits by change of variable problem
I have two limits, to be solved by changing variables, that I'm having difficulty with:

lim x-> 1

x^(1/6) - 1
_________

x-1

I let x^(1/6) = u. When x -> 1, u -> 1, and made the denominator u^6 - 1. However, this still gives a denominator of 0, so I'm not sure if I subbed the values correctly, or if I'm just not seeing how this should be factored. The other one, which I'm unsure how to do, is:

lim x-> 1

x^(1/6) - 1
________

x^(1/3) - 1
• Jul 12th 2007, 11:44 AM
ThePerfectHacker
Quote:

Originally Posted by starswept
I have two limits, to be solved by changing variables, that I'm having difficulty with:

lim x-> 1

x^(1/6) - 1
_________

x-1

$\lim_{x\to 1} \frac{x^{1/6} - 1}{x-1} = \lim_{y\to 1}\frac{y-1}{y^6-1} = \lim_{y\to 1} \frac{y-1}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$
• Jul 12th 2007, 11:46 AM
topsquark
Quote:

Originally Posted by starswept
I have two limits, to be solved by changing variables, that I'm having difficulty with:

lim x-> 1

x^(1/6) - 1
_________

x-1

I let x^(1/6) = u. When x -> 1, u -> 1, and made the denominator u^6 - 1. However, this still gives a denominator of 0, so I'm not sure if I subbed the values correctly, or if I'm just not seeing how this should be factored.

$\lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1}$

Let $u = x^{1/6}$

Then we have
$\lim_{u \to 1} \frac{u - 1}{u^6 - 1}$

$= \lim_{u \to 1} \frac{u - 1}{(u - 1)(u^5 + u^4 + u^3 + u^2 + u + 1)}$

$= \lim_{u \to 1} \frac{1}{u^5 + u^4 + u^3 + u^2 + u + 1} = \frac{1}{6}$

-Dan
• Jul 12th 2007, 11:49 AM
topsquark
Quote:

Originally Posted by starswept
The other one, which I'm unsure how to do, is:

lim x-> 1

x^(1/6) - 1
________

x^(1/3) - 1

$\lim_{x \to 1} \frac{x^{1/6} - 1}{x^{1/3} - 1}$

Let $u = x^{1/6}$

Then you have
$\lim_{u \to 1} \frac{u - 1}{u^2 - 1}$

You should be able to finish this from here.

-Dan
• Jul 12th 2007, 11:51 AM
topsquark
Quote:

Originally Posted by topsquark
$\lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1}$

Let $u = x^{1/6}$

Then we have
$\lim_{u \to 1} \frac{u - 1}{u^6 - 1}$

$= \lim_{u \to 1} \frac{u - 1}{(u - 1)(u^5 + u^4 + u^3 + u^2 + u + 1)}$

$= \lim_{u \to 1} \frac{1}{u^5 + u^4 + u^3 + u^2 + u + 1} = \frac{1}{6}$

-Dan

Or, you know, there's always L'Hopitals rule. (I'm going to duck all the rotten tomatos being thrown at me!)

$\lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{6}x^{-5/6}}{1} = \frac{1}{6}$

all in one line! :)

-Dan
• Jul 12th 2007, 11:54 AM
starswept
Quote:

Originally Posted by topsquark
Or, you know, there's always L'Hopitals rule. (I'm going to duck all the rotten tomatos being thrown at me!)

$\lim_{x \to 1} \frac{x^{1/6} - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{6}x^{-5/6}}{1} = \frac{1}{6}$

all in one line! :)

-Dan

That's a nifty one! Never seen it before :) Thanks!
• Jul 12th 2007, 02:08 PM
topsquark
Quote:

Originally Posted by starswept
That's a nifty one! Never seen it before :) Thanks!

If you haven't yet, you will...

-Dan