# Math Help - Limits Question

1. ## Limits Question

On a practice exam for our exam tomorrow our professor has a problem

lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

He has negative infinity as the correct answer?

2. Originally Posted by KarlosK
On a practice exam for our exam tomorrow our professor has a problem

lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

He has negative infinity as the correct answer?
$\lim_{x \to 3^-} \frac{x^2 (2-x)}{2x^2 -5x -3}=\lim_{x \to 3^-} \frac{x^2 (2-x)}{(2x+1)(x-3)}$

Because x=3 does not make the numerator zero, it is going to be an infinity type of limit. To determine which type, think about the sign of x-3 as x approaches 3 from the left. So for x=2.999, note that x-3 is very close to zero, but still negative. Hence the overall fraction will approach negative infinity.
I hope this helps!

Good luck.

3. Originally Posted by KarlosK
On a practice exam for our exam tomorrow our professor has a problem

lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

He has negative infinity as the correct answer?
To clarify, negative infinity is what the function is approaching as $x\rightarrow 3^-$, but since infinity does not exist, even the one-sided limit does not exist.

Hope that clears it up

Mathemagister