# continuity

• Apr 6th 2006, 01:04 AM
rachael
continuity
5 Consider the function f with rule f(x) {(x^2 - 25 )/ (x-5) if x ≠ 5
a if x = 5

If f is continuous on R, find the value of a.
• Apr 6th 2006, 06:43 AM
ThePerfectHacker
Quote:

Originally Posted by rachael
5 Consider the function f with rule f(x) {(x^2 - 25 )/ (x-5) if x ≠ 5
a if x = 5

If f is continuous on R, find the value of a.

I understand you question as,
$\displaystyle f(x)=\frac{x^2-25}{x-5},x\not =5$ is a real-valued function. It is definitely not countinous on $\displaystyle \mathbb{R}$ because 5 is not in its domain. Thus, you are asking how do define the function at $\displaystyle x=5$ thus it would be countinous? Notice you function is, $\displaystyle \frac{x^2-25}{x-5}=\frac{(x-5)(x+5)}{x-5}$=$\displaystyle x+5$. This a linear function not defined at $\displaystyle x=5$. Thus, we need to redefine it at $\displaystyle x=5$ call that function $\displaystyle g(x)$. We need not do redefine it anywhere else because it is linear and is countinous. By the definition of countinuity we need that,
$\displaystyle \lim_{x\to 5}g(x)=g(5)$ Since, $\displaystyle f$ and $\displaystyle g$ are function that agree at all but one point (at 5) we have that they have the same limit. Thus, $\displaystyle \lim_{x\to 5}f(x)=g(5)$, but $\displaystyle \lim_{x\to 5}f(x)=\lim_{x\to 5}x+5=10$. Thus, we have $\displaystyle g(5)=10$. Thus, the function,
$\displaystyle g(x)=x+5$ is the redefined countinous function of $\displaystyle f(x)=\frac{x^2-25}{x-5}$.