# Math Help - Some Limit Questions

1. ## Some Limit Questions

Having some issues with these questions:

1.) limit as x ---->4 $\frac{x^2-x-12}{2x^2-5x-12}$
I know I can factor the numerator to have: (x-4 (x+3).

2.) limit as x---->infinity $\frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}$

Do I find a common denominator first, subtract the fractions, and take it from there?

3.) limit as x---->2 $\frac{\sqrt{x+2}-2}{x-2}$

I think I use the conjugate here.

Well, that is how I tacked the problem. Will post my new answers shortly if I think of a better way to answer them

TIA!

2. Originally Posted by mvho
Having some issues with these questions:

1.) limit as x ---->4 $\frac{x^2-x-12}{2x^2-5x-12}$
I know I can factor the numerator to have: (x-4 (x+3).
The denominator factors to $(2x+3)(x-4)$. Can you take it from here?

2.) limit as x---->infinity $\frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}$

Do I find a common denominator first, subtract the fractions, and take it from there?
Note that if you multiply the first fraction by $\frac{\frac{1}{x^2}}{\frac{1}{x^2}}$, you get $\frac{1}{1-\frac{1}{x^2}}$; if you multiply the second fraction by $\frac{\frac{1}{x^4}}{\frac{1}{x^4}}$, you get $\frac{\frac{1}{x}+\frac{1}{x^4}}{1+\frac{1}{x^4}}$

Thus, $\lim_{x\to\infty}\frac{x^2}{x^2-1}-\frac{x^3+1}{x^4+1}=\lim_{x\to\infty}\frac{1}{1-\frac{1}{x^2}}-\frac{\frac{1}{x}+\frac{1}{x^4}}{1+\frac{1}{x^4}}$

Also note that $\lim_{x\to\infty}\frac{1}{x^n}=0$ whenever $n\geq1$.

Can you take it from here?

3.) limit as x---->2 $\frac{\sqrt{x+2}-2}{x-2}$

I think I use the conjugate here.

Well, that is how I tacked the problem. Will post my new answers shortly if I think of a better way to answer them

TIA!
Yes, you would use conjugate here. How did you do with this one?

I hope this all made sense to you...

3. 1) Factor the denominator.

$2x^2-5x-12=2x^2-8x+3x-12=2x(x-4)+3(x-4)=(x-4)(2x+3)$

2)In this case you can use the limits of the rational functions when $x\to\infty$:

- if both nomerator and denominator have the same degree then the limit is $\frac{a_n}{b_n}$, where $a_n,b_n$ are the dominant coefficients.

-if the denominator has a degree greater then numerator, then the limit is 0.

3) $\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}=\lim_{x\to 2}\frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{x-2}=\lim_{x\to 2}\frac{x-2}{(x-2)(\sqrt{x+2}+2)}$

4. Thanks guys! Chris, that made a lot of sense. I just suck hardcore at factoring. When I saw that question, I knew the denominator would screw me over.

For the 2nd question, since x is going to infinity essentially a small number divided by a really big number will approach 0.

For the 3rd one, looks like I forgot to multiply -2 x 2, but I worked it out. I got $\frac{1}{4}$ as my answer.

As for the 1st one, I am really not good at factoring as embarrassed to say I don't see how you factored it. Is there some kind of method?

Like, how did you go from:

$2x^2-5x-12$ to $2x^2-8x+3x-12$

From there, I can see how you factor it easily. Just can't make that initial step

5. Originally Posted by red_dog

3) $\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}={\color{red}{\lim_{x\to 2}\frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{x-2}}}=\lim_{x\to 2}\frac{x-2}{(x-2)(\sqrt{x+2}+2)}$

Do you mean this?

$\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2} - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {\sqrt {x + 2} - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}}.$

or

$\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2} - 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2} - 2}}{{x + 2 - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {x + 2} - 2}}{{\left( {\sqrt {x + 2} - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}}.$

6. Originally Posted by mvho
I am really not good at factoring as embarrassed to say I don't see how you factored it. Is there some kind of method?
Yes.