Finding where a function is discontinous, need some help

[Mathamateur]

I'm having a little trouble with this problem. I think I might have the answer but I'm not positive.

Directions: Find all x values where the function is discontinuous. For each such value, give f(a) and lim_x->a f(x).

Problem 1) f(x) = (x^2 - 9) / (x + 3)

I got -3 for the x value where the function is discontinuous and then had undefined for f(a) and -6 for the limit part. Am I right, or at least close? I wasn't quite sure what they were asking for.

The next one I am just completely lost on. (Same directions as problem 1.)

2) f(x) = 2x^2 - 5x - 3

Thanks in advance

 

[tkhunny]

This post makes very little sense. Unfortunately, you did not actually show your work, so it is very difficult to guide you on a better path. "I got" followed by "and then had" hardly communicates what it is you are doing.

For \(\displaystyle f(x) = \frac{x^{2}-9}{x+3}\), you have it correct that it is discontinuous at x = -3. After that, you totally lost me. What has 'a' to do with anything? What is '-6'? What limit?

On the second one, that's just a polynomial, it's continuous everywhere. They always are.

 

[stapel]

How does "a" relate to f(x)? Is this the x-value at which f(x) is discontinuous?

1) Since (x² - 9) / (x + 3) = [(x + 3)(x - 3)] / (x + 3) = x - 3 except for x = -3, then, yes, the limit as x -> -3 of f(x) is [-3] - 3 = -6. (This assumes that "a" is defined according to my guess above.) The original function, of course, is not defined at this x-value.

2) This is a polynomial. Do polynomials have any "problem" points?

Eliz.