# Math Help - Evaluating Limits

1. ## Evaluating Limits

Hi, I'm not sure how to evaluate the following limits.

1. lim (x^3+x^2-8x-12)/(x+2)
x->-2

so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).

2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??

2. Hi skeske1234

If you know 3 is the facor of the numerator, can you factorize (x^3+x^2-8x-12) ?

3. Originally Posted by songoku
Hi skeske1234

If you know 3 is the facor of the numerator, can you factorize (x^3+x^2-8x-12) ?
yes.. well I'm not sure If this is right.. am i supposed to put x=3 in numerator? so I will get 0 on the numerator? or am I supposed to put in (x-3) for x?

if I put in 0, what do I do next?

4. Originally Posted by skeske1234
Hi, I'm not sure how to evaluate the following limits.

1. lim (x^3+x^2-8x-12)/(x+2)
x->-2

so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).
The zeros of the numerator are not of primary importance here. The problem is that the denominator has a zero at the very point where we want to take the limit! So if you first check (by polynomial division) whether the factor $x+2$ occurs in the numerator you find that
$\begin{array}{lcl}
\lim\limits_{x\to-2}\frac{x^3+x^2-8x-12}{x+2}&=&\lim\limits_{x\to -2}\frac{(x+2)(x^2-x-6)}{x+2}\\
&=&\lim\limits_{x\to -2}(x^2-x-6)\\
&=&(-2)^2-(-2)-6\\
&=&0
\end{array}$

2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??

Remember $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots + ab^{n-2}+b^{n-1})$? Now substitute $y$ for $x^{1/3}$ (and therefore $y^3$ for $x$), and you find that

$\begin{array}{lcl}
\lim\limits_{x\to 27}\frac{27-x}{x^{1/3}-3} &=&\lim\limits_{y\to 3}\frac{3^3-y^3}{y-3}\\
&=&\lim\limits_{y\to 3}\frac{(3-y)(3^2+3y+y^2)}{y-3}\\
&=&\lim\limits_{y\to 3}[-(9+3y+y^2)]\\
&=& -(9+3\cdot 3+3^2)\\
&=& -27
\end{array}$

5. Originally Posted by skeske1234
Hi, I'm not sure how to evaluate the following limits.

1. lim (x^3+x^2-8x-12)/(x+2)
x->-2

so I know that I have to use the factor theorem for the numerator, and I noticed that 3 is a factor of the numerator. so I'm not sure how to evalute this limit or proceed to the next step with this in mind (using the factor theorem).
Do you mean that "3" is a factor of the numerator or that "x- 3" is a factor?
In any case, that is not really relevant. What is very relevant is that x+2, the denominator, goes to 0 as x goes to -2 so "x+2" had better be a factor! You can show that it is, and find the factorization by dividing $x^3+ x^2- 8x- 12$ by x+ 2, either by direct division or by "synthetic" division. You could also "work" it out by thinking "if x+2 is a factor, then it must be something like $x^3+ x^2- 8x- 12= (x+ 2)(x^2+ ?x- 6)$ Multiply out the left side and see what "?" must be.

(If you know x+ 2 must be a factor and you also know that x- 3 is a factor, then it is easy to see that [tex]x^3+ x^2- 8x- 12= (x+ 2)(x- 3)(x- ?) and since -12= 2(-3)(2), what is "?" ?

2. lim(27-x)/(x^(1/3)-3)
x->27
for this one, i am not sure if i have to change the variables and if so, what i have to change it to??
Again, you need to be able to cancel that $x^{1/3}- 3$ in the denominator. Fortunately, we can precisely because 27 also makes the numerator 0. Remember that $a^3- b^3= (a- b)(a^2+ ab+ b^2)$ with $a^3= 27$ and $b^3= x$.