1. ## Limits Help

Limit as x goes to negative infinity

$\displaystyle (3x^2 + 1)^-2$

(Last part is raised to the -2 power not minus 2)

2. Originally Posted by NRS11
Limit as x goes to negative infinity

$\displaystyle (3x^2 + 1)^-2$

(Last part is raised to the -2 power not minus 2)
like this?

$\displaystyle \lim_{x\to\infty^-}\left(3x^2+1\right)^{-2}$

or

$\displaystyle \lim_{x\to\infty^-}\frac{1}{\left(3x^2+1\right)^{2}}$

3. The top one that you mentioned. It is NOT the fraction one.

4. they are both the same thing

the 2nd is easier to work with

5. The key is to know that for $\displaystyle n>0$ is $\displaystyle \frac1{t^n}\to0$ as $\displaystyle t\to+\infty.$

Now, in your limit put $\displaystyle t=-x$ so the limit equals $\displaystyle \lim_{t\to+\infty}\frac1{(3t^2+1)^2},$ so for large $\displaystyle t$ is $\displaystyle \frac{1}{16t^{4}}<\frac{1}{\left( 3t^{2}+1 \right)^{2}}<\frac{1}{9t^{4}},$ and the squeeze theorem finishes the problem.

6. $\displaystyle \frac{1}{(3x^2+1)^2}$

is simply a bell-shaped curve with a maximum of 1 at x=0.

As x approaches + or -infinity, f(x) rapidly goes to zero.