Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've done so many limit problems in calculus lately, but I can't wrap my mind around how to simplify this one in order to solve it:

$$ \lim_{x\rightarrow 2} \dfrac{x^3-8}{x^2-x-2} $$

I understand the $x^3-8$ factors down to $(x-2)(x^2+2x+4)$, but that still leaves us with $$ \lim_{x\rightarrow 2} \dfrac{(x-2)(x^2+2x+4)}{x^2-x-2}, $$ which I can't seem to find a way to simplify so that the denominator is not equal to 0.

In case anyone figures out themselves, the answer is 4 (I was given the answer - this is on a review sheet for an upcoming exam). Also, I tagged this as homework, even though it is not technically homework.

So if anyone could help point me in the right direction here, that would be very helpful.

share|improve this question
1  
The denominator factors as $x^2-x-2 = (x-2)(x+1)$. Does that help? –  Arturo Magidin Oct 8 '11 at 20:35
1  
If you plug $2$ into a polynomial and get $0$, then $x-2$ is one of its factors. That's worth knowing. And if you plug $2$ in and get something other than $0$, then you won't get a $0$ in the denominator in a case like this, so then you could just plug $2$ into the whole expression and that's the limit. –  Michael Hardy Oct 8 '11 at 21:20

3 Answers 3

up vote 2 down vote accepted

Hint: Note that $x^2-x-2=(x-2)(x+1)$.

share|improve this answer
1  
And theres my answer.. I didn't even think that the denominator was factorable! How could I have missed that. Thank you - I will approve your answer when the time limit for which I can't approve any answer is up. –  Mike Gates Oct 8 '11 at 20:36
10  
@Mike: If $p(x)$ is a polynomial, and $p(a)=0$, then $p(x)$ can always be factored as $p(x)=(x-a)q(x)$ for some polynomial $q(x)$. This is the Factor Theorem, so with these kinds of limits (a rational function that evaluates to $\frac{0}{0}$), you can always factor both the numerator and the denominator, cancel the factor, and try again. –  Arturo Magidin Oct 8 '11 at 20:37
    
Oh okay. That's sensible. –  Mike Gates Oct 8 '11 at 20:44

You can use the L`Hospital rule to find the answer. L´Hospital Rule

share|improve this answer
    
Why this late answer to a question with an accepted answer? And it's not like your answer is much better than the previous answers either. –  TMM Mar 30 '13 at 15:42

$$\lim_{x \to 2}\dfrac{x^3-8}{x^2-x-2}=\lim_{x \to 2}\dfrac{(x^3-8)'}{(x^2-x-2)'}=...$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.