1. ## Basic Limits Help

Reviewing for the final exam, forgot a couple simple rules.

To solve a limit algebraically

1. lim x -> -5 (x^2 +3x - 10) / (x + 5)

If you plug in -5 it is 0/0 so I factored the top to (x+5)(x-2) but the bottom is still 0. I wrote no limit for the answer to this problem.

2. lim x-> 3- (x)/(x^2 - 9)

Again the limit is 0/0. I factored the bottom to be (x+3)(x-3) but plugging in 3 you wind up getting a number over 0 again. Again I think the answer is no limit.

Am I forgetting something here?

2. Originally Posted by KarlosK
Reviewing for the final exam, forgot a couple simple rules.

To solve a limit algebraically

1. lim x -> -5 (x^2 +3x - 10) / (x + 5)

If you plug in -5 it is 0/0 so I factored the top to (x+5)(x-2) but the bottom is still 0. I wrote no limit for the answer to this problem.

2. lim x-> 3- (x)/(x^2 - 9)

Again the limit is 0/0. I factored the bottom to be (x+3)(x-3) but plugging in 3 you wind up getting a number over 0 again. Again I think the answer is no limit.

Am I forgetting something here?
1:

$\displaystyle \lim_{ x \to -5 } \frac{ x^2 +3x - 10 }{ x + 5 } = \lim_{ x \to -5 } \frac{ (x+ 5)(x-2) }{ x + 5 } = \lim_{ x \to -5 } (x-2) = -7$

2:

$\displaystyle \lim_{x \to 3^- } \frac{x}{x^2 - 9}$

We can do partial fractions here, but it is sufficient to note that 3 is being approached from the negative side (i.e. 2.99, 2.999, 2.9999, etc) and that makes the denominator a small negative quantity.

Thus, a static number divided by an infinately small negative number is negative infinity.

$\displaystyle \lim_{x \to 3^- } \frac{x}{x^2 - 9} = - \infty$