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.
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.
Because is...
$\displaystyle f(x)= \frac{|x-2|}{x^{2}- 5 x + 6} = \left\{\begin{array}{ll} \frac{1}{x-3} ,\,\, x >2\\{}\\ \frac{1}{3-x} ,\,\, x<2\end{array}\right.$ (1)
... is...
$\displaystyle \lim_{x \rightarrow 2 +} f(x)= -1$
$\displaystyle \lim_{x \rightarrow 2 -} f(x)= 1$ (2)
... so that f(x) isn't continous for $\displaystyle x=2$ no matter which is the value You assign to it in that point...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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, $\displaystyle \frac{|x- 2|}{x^2- 5x+ 6}$ is not defined at x= 2.
Fortunately, the definition of "$\displaystyle \lim_{x\to a} f(x)= L$" says that "Given $\displaystyle \epsilon> 0$, there exist $\displaystyle \delta> 0$ such that if 0< $\displaystyle |x- a|< \delta$, then $\displaystyle |f(x)- L|< \epsilon$". 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.