I don't know how to format, sorry...
"Find the limit, if it exists"
Limit (as x approaches 2):
I used the quadratic formula to try to find values for the numerator, but one ends up with 1 plus or minus the square root of -23 divided by two. Obviously, negative numbers don't have real square roots, so is it safe to say there is no limit to this equation? Thanks...
Not quite, satx,
if they were converging on the same infinity,
then both branches would be shooting off in the same direction,
once again never meeting.
In a situation like this, we check to see if the denominator is
a factor of the numerator.
If it was, we could say for all x except 2.
Then that situation is described as having a "hole" in the graph at x=2.
If the numerator is a quadratic and the denominator linear,
the graph would be indistinguishable from a straight line.
Written as a fraction of course, we'd have to exclude 2 from the domain
if the denominator contains (x-2).
In that case the limit may be evaluated as the graph approaches f(2)
from both sides, since if x is not 2, then
hence
has a "hole" at x=2
though the graph is indistinguishable from x-3.
Hence f(2) would be -1 if the graph was x-3,
therefore -1 is the limit as x approaches 2 for
We cannot "see" the hole, of course.
For
the denominator is not a factor of the numerator.
Another way to express the function is
If x is 2, then again cannot be evaluated
since approaches infinity as x approaches 2.