# Help in evaluating limit

1. Aug 31, 2007

### f(x)

1. The problem statement, all variables and given/known data
Evaluate: $$\lim_{x \rightarrow \infty} ( \frac {x^3}{(3x)^2-4}-\frac{x^2}{3x+2} )$$

2. The attempt at a solution

First i took x^2/3x+2 common, and then substituted y=1/x , y->0.
simplifying, i get $$\frac{-2}{9y}$$...i am unable to eliminate all the y's.
The answer given at the back of the book is 2/9.
Plz help me figure this out.
&&Thx

Last edited: Aug 31, 2007
2. Aug 31, 2007

### learningphysics

Are you sure that's exactly the question? The limit approaches minus infinity...

3. Aug 31, 2007

### P.O.L.A.R

There is a theorm for this because what you have is an $$\infty$$/$$\infty$$ correct? (L'Hopital)
Note: If this is a calc one qustion ignore this.

Last edited: Aug 31, 2007
4. Aug 31, 2007

### f(x)

yeah, thats why i am confused

Say for the 1st term, i divide and multiply by x^3, the Dr. becomes 0 ?

5. Aug 31, 2007

### learningphysics

Are you sure the question isn't:

$$\lim_{x \rightarrow \infty} ( \frac {3x^3}{(3x)^2-4}-\frac{x^2}{3x+2} )$$

6. Aug 31, 2007

### rock.freak667

i keep getting $$\infty$$ not 2/9

7. Sep 1, 2007

### d_leet

Find a common denominator, then add the fractions together.

8. Sep 1, 2007

### f(x)

I checked it again, but the question is exactly as I put it.

9. Sep 1, 2007

### malawi_glenn

(3x)^2 - 4 = 9x^2 - 4 = (3x + 2)(3x - 2)

use this to simply your function (i.e multply your second term with (3x -2) in both nominator and denominator)

Then the x^3 term will vanish

Also, there must be a misprint as "learningphysics" has noticed

Last edited: Sep 1, 2007
10. Sep 1, 2007

### learningphysics

I'm guessing it's a misprint or something that left out the 3... because the limit of the function I just posted is 2/9.