Is a vertical asymptote possible at x=-6 if f(-6) = 3?

[Loppyfoot]

Homework Statement

If f(-6) = 3, is it possible for there to be a vertical asymptote at x=-6?
The limit as x approaches -6 is positive infinity, but there f(-6) is defined by the value of -3. Or, since VAs refer to the line, the fact that f(-6) doesn't matter?

 

[Mark44]

How f is defined at x = -6 doesn't matter.

Here is an example function that has the behavior you describe.

f(x) = 1/(x + 6), if x != -6
f(x) = 3, if x = -6

This function has a vert. asymptote at x = -6, even though f(-6) is defined.

 

[Loppyfoot]

Thanks!
Another question:
For the piecewise,
f(x){-x+1 , 0<=x<1
{1, 1<=x<2
What is the limit as x approaches 1?

Do I do?
-x+1 = 1
-x=0
x=0?
So 0+1 = 1
and 1
So the limit is 1? Or does it not exist?

 

[Mark44]

For a limit to exist, both one-sided limits must exist and must be equal.
So you need to find these limits:
\lim_{x \rightarrow 1^+}f(x)
\lim_{x \rightarrow 1^-}f(x)
If both exist and are equal, then
\lim_{x \rightarrow 1}f(x)
exists and is equal to the common value.