# Math Help - Stuck on Limit Problem

1. ## Stuck on Limit Problem

Hi, I don't know how to approach this...I'm stuck ... I tried using the conjugate to find the limit and solved for $x$, but it still doesn't work... it always give me zero (unless I did it wrong). Can you show me how to solve this?

$

\lim_{x \to 4} \frac{x^3-64}{x-4}

$

2. If you plug four in, you get 0/0, so you need to apply L'Hopital's Rule and take the derivative of the numerator and the derivative of the denominator (separately, not using the quotient rule). This gives you the limit as x approaches 4 of 3 x^2 or 48.

Edit: Hope that helped! Also, what did you use to get the expression as a nice little picture with all the symbols? I'm new to this.

3. Originally Posted by kdl00
Hi, I don't know how to approach this...I'm stuck ... I tried using the conjugate to find the limit and solved for $x$, but it still doesn't work... it always give me zero (unless I did it wrong). Can you show me how to solve this?

$

\lim_{x \to 4} \frac{x^3-64}{x-4}

$
$x^3-64=(x-4)(x^2+4x+16)$.

4. Originally Posted by kdl00
Hi, I don't know how to approach this...I'm stuck ... I tried using the conjugate to find the limit and solved for $x$, but it still doesn't work... it always give me zero (unless I did it wrong). Can you show me how to solve this?

$

\lim_{x \to 4} \frac{x^3-64}{x-4}

$

Reduce $

\lim_{x \to 4} \frac{x^3-64}{x-4}

$
to

(x-4)(x^2+4x+16) / (x-4)

and evaluate the expression (x^2+4x+16) as limit tends to 4

5. That way works too! Another method would be long division if you can't remember the formula for those darned cubics.

6. Well, there is another method.
take $f(x)=x^3$ then $f(4)=64$.
By the limit definition of the derivative:

$\lim_{x\to 4} \frac{x^3-64}{x-4}=f'(4)=3(4)^2=48$

7. Originally Posted by General
$x^3-64=(x-4)(x^2+4x+16)$.
lol that was simpler then i thought... thank you!