# Thread: This equation doesn't have a limit, does it?

1. ## This equation doesn't have a limit, does it?

I don't know how to format, sorry...

"Find the limit, if it exists"

Limit (as x approaches 2): $(x^2 - x +6)/(x-2)$

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...

2. Originally Posted by satx
I don't know how to format, sorry...

"Find the limit, if it exists"

Limit (as x approaches 2): (x^2 - x +6)/(x-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...
Yes satx,

here is a sketch of it

3. Originally Posted by satx
I don't know how to format, sorry...
Why not learn to post in symbols? You can use LaTeX tags.
$$\lim _{x \to 2} \frac{{x^2 - x + 6}}{{x - 2}}$$ gives $\lim _{x \to 2} \frac{{x^2 - x + 6}}{{x - 2}}$.

4. Originally Posted by Archie Meade
Yes satx,

here is a sketch of it
So because 2+ and 2- are converging on opposite infinities, there's no limit? If they were to converge on the same infinity, there would be a limit, right?

5. 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 $\frac{x-2}{x-2}=1,$ 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

$\frac{x-2}{x-2}=1$

hence

$\frac{x^2-5x+6}{x-2}=\frac{(x-2)(x-3)}{x-2}$

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 $\frac{x^2-5x+6}{x-2}$

We cannot "see" the hole, of course.

For $\frac{x^2-x+6}{x-2}$

the denominator is not a factor of the numerator.

Another way to express the function is $f(x)=(x+1)+\frac{8}{x-2}$

If x is 2, then $\frac{8}{x-2}$ again cannot be evaluated

since $\frac{8}{x-2}$ approaches infinity as x approaches 2.