Question

lim x^2/x^2-4
x-> -2^-

Answer 1

\[\lim_{x \rightarrow -2^-}\frac{ x^2 }{ x^2 -4}\]

Answer 2

Hey melissa :)

So if we plug -2 directly in we get the indeterminate form: \(\Large\rm \frac{4}{0}\) yah?

Answer 3

Yeah, does that mean that it's continuous and that the limit is infinity?

Answer 4

So we have some type of discontinuity at -2.
It's not a removable discontinuity since we aren't able to cancel out the (x+2) factor.
So we have an asymptote at x=-2.

We need to figure out if it's blowing up to positive infinity from the left side,
or down to negative infinity from that side.

Answer 5

We start by factoring the denominator,\[\Large\rm \lim_{x\to-2^{-}}\frac{ x^2 }{(x-2)(x+2)}\]To determine which direction this is going in from the left side of -2 we need only concern ourselves with the signs.
Notice the numerator will always be positive.

Answer 6

\[\Large\rm \lim_{x\to-2^{-}}\frac{+}{(x-2)(x+2)}\]How bout the bottom?
If x is less than negative 2, that means it's MORE NEGATIVE, yah?
So let's use a sample number that's close by,\[\Large\rm \frac{+}{(-2.0001-2)(-2.0001+2)}\]

Answer 7

Is our denominator going to be positive or negative? :o

Answer 8

Positive?

Answer 9

Mmm yah good!
Looks like we're getting something like this,\[\Large\rm \frac{+}{(-)(-)}\]

Answer 10

We already determined that it would be approaching positive or negative infinity, it's asymptotic here at -2.

And now we further determined that when we approach -2 from the LEFT SIDE, we blow up towards positive infinity! :)

Answer 11

(Since the sign is positive overall).

Answer 12

I know, I know, math is crazy :U

Answer 13

calm down melissa :U

Answer 14

No, I understand it haha. I already had the answer, I was just trying to make sure that I had it right.

Answer 15

It's just important that you understand that you have to be thorough when they give you that little (from the left or from the right) thing.

If you were approaching -2 from the RIGHT SIDE, you would go down towards negative infinity.

Answer 16

Yeah, that's the part that I didn't know how to show. I know the answer by looking at it but I don't know how to actually write it down in paper besides just writing down the answer, do you understand?

Answer 17

So the answer would be lim --> -2^- x^2/x^2-4 = +infinity with an asymptote at x=-2 ?

Answer 18

I don't know if my method would be considered "rigorous" but it works.
Choose a point on the right side of -2,\[\Large\rm \frac{+}{(-1.9999-2)(-1.9999+2)}\]And again, figure out the overall sign.

Answer 19

Yes, it's approaching infinity :)
Depending on who your teacher is....
it might drive him crazy if you put an equal sign next to an infinity, they really hate that.

It might be better to say:
\(\Large\rm \frac{x^2}{x^2-4}\to\infty\quad\) as \(\Large\rm \quad x\to-2^{-}\)
Maybe your teacher isn't that picky though hehe.

Answer 20

My teachers always have it written the way I did it so I'd rather not change it and have them fuss about it haha. Thank you so much, though.

Answer 21

ok cool c:

Answer 22

Mind helping with one more? It's similar to this one.

Answer 23

k

Answer 24

\[\lim_{x \rightarrow 2+}\frac{ x }{ \sqrt{x^2-4} }\]

Answer 25

I did direct substitution and ended up with 2/0 but I'm not sure where to go from there.

Answer 26

So that tells us that we have some type of discontinuity at x=2.
Either removable or asymptotic.

It's a removable discontinuity if we can cancel out the problem somehow.

Answer 27

But that's not the case here.
(x-2)(x+2)
We need to cancel out the x-2, but we are unable.
So our discontinuity is asymptotic.

Answer 28

Then it's a non removable one?

Answer 29

Yah, so it's just like the last problem, it's either blowing up or down.

Answer 30

Let's plug in a number that's on the right side of positive 2, and see what it's looking like,\[\Large\rm \frac{2.01}{\sqrt{(2.01)^2-4}}\]

Answer 31

Positive?

Answer 32

Oh notice that the top is always positive around positive 2.
So again, we only care about the bottom.

And notice here, we don't have the option for it to go down to negative infinity. It's either blowing up towards positive infinity, or it doesn't exist (because we can't take the root of a negative).

Answer 33

Mmm yah, positive! :)

Answer 34

Thanks a lot again, haha. Do you mind looking at my other question and seeing if you know what they're talking about? None of my examples are like that one so I'm confused.

Answer 35

sure :o i can spare a few minutes.