Different kind of Limit Problem

**polymerase** · November 5th 2007, 12:33 PM

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

**Jhevon** · November 5th 2007, 12:46 PM

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?

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

**polymerase** · November 5th 2007, 12:50 PM

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

**Jhevon** · November 5th 2007, 01:01 PM

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?

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

what about now?

**polymerase** · November 5th 2007, 01:15 PM

Originally Posted by Jhevon

do you not see the limit we want in there?

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

what about now?

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

**Jhevon** · November 5th 2007, 01:20 PM

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

**polymerase** · November 5th 2007, 01:24 PM

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

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