Solving limits by change of variable

• Jul 11th 2007, 06:01 PM
starswept
Solving limits by change of variable
I learned today that one of the methods for solving a limit is by change of variable. However, I'm confused as to how it works. I have the following question as an example:

Evaluate the limit by change of variable.

lim x->8

3√x - 2
______
x-8

I started by saying let 3√x = y. So the numerator becomes y-2. However, I'm uncertain how to change the denominator when solving by change of variable , and how to determine what the new lim y-> is (are there general rules for these?). Thank you so much for any help, as I have a test on this on Friday.
• Jul 11th 2007, 06:05 PM
Jhevon
Quote:

Originally Posted by starswept
I learned today that one of the methods for solving a limit is by change of variable. However, I'm confused as to how it works. I have the following question as an example:

Evaluate the limit by change of variable.

lim x->8

3√x - 2
______
x-8

I started by saying let 3√x = y. So the numerator becomes y-2. However, I'm uncertain how to change the denominator when solving by change of variable , and how to determine what the new lim y-> is (are there general rules for these?). Thank you so much for any help, as I have a test on this on Friday.

just to clarify, by 3√x do you mean 3 times the square root of x or the cube root of x?
• Jul 11th 2007, 06:06 PM
starswept
Quote:

Originally Posted by Jhevon
just to clarify, by 3√x do you mean 3 times the square root of x or the cube root of x?

Sorry about that, I mean the cube root of x :)
• Jul 11th 2007, 06:43 PM
Jhevon
Quote:

Originally Posted by starswept
Sorry about that, I mean the cube root of x :)

i am not familiar with this change of variable method you refer to. i'd do this problem by realizing the denominator is the difference of two cubes, of which one of the factors will cancel with a term in the top
• Jul 11th 2007, 10:29 PM
red_dog
Let $\displaystyle \sqrt[3]{x}=y\Rightarrow x=y^3$.
If $\displaystyle x\to 8$ then $\displaystyle y\to 2$.
So $\displaystyle \displaystyle \lim_{x\to 8}\frac{\sqrt[3]{x}-2}{x-8}=\lim_{y\to 2}\frac{y-2}{y^3-8}=\lim_{y\to 2}\frac{y-2}{(y-2)(y^2+2y+4)}=$
$\displaystyle \displaystyle =\lim_{y\to 2}\frac{1}{y^2+2y+4}=\frac{1}{12}$