# Continuity

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• May 1st 2010, 10:36 PM
Monster32432421
Continuity
The function Abs[x - 2]/(x^2 - 5x + 6) is not defined for x=2. Is there a value that could be given for f(2) that would make the function continuous at 2?
Give reasons for your answer.
• May 1st 2010, 10:41 PM
maddas
You dropped an x on the denominator I think. Try factoring.
• May 1st 2010, 10:50 PM
Monster32432421
Quote:

Originally Posted by maddas
You dropped an x on the denominator I think. Try factoring.

yea my bad. You factorise it and simplify to get 1/(x-3) but then what would that mean?
• May 1st 2010, 10:52 PM
maddas
Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....
• May 1st 2010, 11:04 PM
Monster32432421
Quote:

Originally Posted by maddas
Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....

|x-2|/(x-3)(x-2) = 1/(x-3)
or cant i cancel them cause its abs?
• May 1st 2010, 11:06 PM
maddas
You tell me. Are both sides of that equation actually equal for all x? What about x=1?
• May 1st 2010, 11:06 PM
chisigma
Because is...

$f(x)= \frac{|x-2|}{x^{2}- 5 x + 6} = \left\{\begin{array}{ll} \frac{1}{x-3} ,\,\, x >2\\{}\\ \frac{1}{3-x} ,\,\, x<2\end{array}\right.$ (1)

... is...

$\lim_{x \rightarrow 2 +} f(x)= -1$

$\lim_{x \rightarrow 2 -} f(x)= 1$ (2)

... so that f(x) isn't continous for $x=2$ no matter which is the value You assign to it in that point...

Kind regards

$\chi$ $\sigma$
• May 1st 2010, 11:08 PM
Monster32432421
^
Thank you

Quote:

Originally Posted by maddas
You tell me. Are both sides of that equation actually equal for all x? What about x=1?

time waster...
• May 1st 2010, 11:10 PM
maddas
Then they're not equal. Anyway, chisigma posted a solution now.

edit: yea, cause its such a waste of time to know that 1/2 is not -1/2.
• May 2nd 2010, 03:57 AM
HallsofIvy
Maddas' point when he wrote "You tell me. Are both sides of that equation actually equal for all x?" was that you cannot cancel "x- 2" because at x= 2, that would be dividing by 0.

[tex]\frac{(x- 3)(x- 2)}{x- 2}= x- 3[/math for all x except x= 2. That's why the original function, $\frac{|x- 2|}{x^2- 5x+ 6}$ is not defined at x= 2.

Fortunately, the definition of " $\lim_{x\to a} f(x)= L$" says that "Given $\epsilon> 0$, there exist $\delta> 0$ such that if 0< $|x- a|< \delta$, then $|f(x)- L|< \epsilon$". That "0< |x- a|" (which many people forget to write) means that the value of f(x) at x= a has no effect whatsoever on the lim as x goes to a.

Although [tex]\frac{(x- 3)(x- 2)}{x- 2}[tex] is not defined at x= 2, it has a limit there and you can make the function continuous by defining f(2) to be that limit.