# Continuity Question

• Aug 28th 2007, 11:47 PM
Wyau
Continuity Question
Sorry about not being able to properly format these questions so that they're easier to read. I'm new here, and I don't really know how to use it yet.

I've been trying to solve this problem for quite some time now, but I still haven't gotten a clear grasp on the limits. It says "Given the two functions f and h suck that f(x) = x^3 - 3x^2 - 4x + 12 and

h(x) {p, for x = 3
f(x)/(x-3), for x is not equal to 3

Find the value of p so that the function h is continuous at x = 3"

Am I supposed to find the limits as it comes from the right and left and then make that the value of p? Any help would be appreciated, as I'm getting very frustrated with all these continuity questions.

• Aug 28th 2007, 11:52 PM
Rebesques
Yeap. In general, that is. Here you don't even need the two-sided limits:

$\lim_{x\rightarrow3}h(x)=\lim_{x\rightarrow3}\frac {f(x)}{x-3}=\lim_{x\rightarrow3}\frac{x^2(x-3)-4(x-3)}{x-3}$, and you take it from here.
• Aug 29th 2007, 01:38 PM
CaptainBlack
Quote:

Originally Posted by Wyau
Sorry about not being able to properly format these questions so that they're easier to read. I'm new here, and I don't really know how to use it yet.

I've been trying to solve this problem for quite some time now, but I still haven't gotten a clear grasp on the limits. It says "Given the two functions f and h suck that f(x) = x^3 - 3x^2 - 4x + 12 and

h(x) {p, for x = 3
f(x)/(x-3), for x is not equal to 3

Find the value of p so that the function h is continuous at x = 3"

Am I supposed to find the limits as it comes from the right and left and then make that the value of p? Any help would be appreciated, as I'm getting very frustrated with all these continuity questions.

When $x \ne 3$

$
h(x)=\frac{x^3 - 3x^2 - 4x + 12}{x-3}=x^2-4
$

but as $x^2-4$ is continuous on $\bold{R}$ for $h(x)$ to be continuous at $x=3$ $p$ must equal $3^2-4 = 5$

RonL
• Aug 29th 2007, 02:35 PM
Wyau
Thanks for your quick responses, guys. It's helped me a lot on my last quiz.