# Different kind of Limit Problem

• Nov 5th 2007, 12:33 PM
polymerase
Different kind of Limit Problem
Suppost that f is a differentiable function such that $f'(8)=2$. Without using L'Hospital's Rule, find the value of $\displaystyle\lim_{x\to 8}\frac{f(x)-f(8)}{x^{\frac{1}{3}}-2}$
• Nov 5th 2007, 12:46 PM
Jhevon
Quote:

Originally Posted by polymerase
Suppost that f is a differentiable function such that $f'(8)=2$. Without using L'Hospital's Rule, find the value of $\displaystyle\lim_{x\to 8}\frac{f(x)-f(8)}{x^{\frac{1}{3}}-2}$

does this help?

$f'(8) = 2 = \lim_{x \to 8} \frac {f(x) - f(8)}{x - 8} = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) }$
• Nov 5th 2007, 12:50 PM
polymerase
Quote:

Originally Posted by Jhevon
does this help?

$f'(8) = 2 = \lim_{x \to 8} \frac {f(x) - f(8)}{x - 8} = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) }$

No not really, I still don't understand how to solve it. I get how you factored....thats about it
• Nov 5th 2007, 01:01 PM
Jhevon
Quote:

Originally Posted by polymerase
No not really, I still don't understand how to solve it. I get how you factored....thats about it

do you not see the limit we want in there?

$2 = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) } = \lim_{x \to 8} \frac {f(x) - f(8)}{x^{\frac 13} - 2} \cdot \frac 1{x^{\frac 23} + 2x^{\frac 13} + 4}$

• Nov 5th 2007, 01:15 PM
polymerase
Quote:

Originally Posted by Jhevon
do you not see the limit we want in there?

$2 = \lim_{x \to 8} \frac {f(x) - f(8)}{\left( x^{\frac 13} - 2 \right) \left( x^{\frac 23} + 2x^{\frac 13} + 4 \right) } = \lim_{x \to 8} \frac {f(x) - f(8)}{x^{\frac 13} - 2} \cdot \frac 1{x^{\frac 23} + 2x^{\frac 13} + 4}$

Ahhh i get it...so the answer is 24 right? Ok back track a little...how did you know that 2 = the limit?
• Nov 5th 2007, 01:20 PM
Jhevon
Quote:

Originally Posted by polymerase
Ahhh i get it...so the answer is 24 right? Ok back track a little...how did you know that 2 = the limit?

yes, it's 24.

we were told f'(8) = 2. i used the limit definition of the derivative
• Nov 5th 2007, 01:24 PM
polymerase
Quote:

Originally Posted by Jhevon
yes, it's 24.

we were told f'(8) = 2. i used the limit definition of the derivative

Ahhhhh....i get it completely now. Thanks