# Having a Limit crisis

## Homework Statement:

lim(x->-2) (x^2 +2x) /(1-abs(x+3))

## Homework Equations:

setting the limit we get 2.
but plugging -2 you clearly get a 0/0 answer. which one is correct?
is x=-2 an asymptote?

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WWGD
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But this is not necessarily a contradiction. This is needed for continuity but not for the existence of a limit.

PeroK
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Homework Statement: lim(x->-2) (x^2 +2x) /(1-abs(x+3))
Homework Equations: setting the limit we get 2.

but plugging 2 you clearly get a 0/0 answer. which one is correct?
is x=-2 an asymptote?
The limit is of the form $0/0$ which is why it's non trivial to evaluate.

It's not an asymptote. It's simply the limit at $x = -2$.

Remember - the limit as a function approaches c is a different concept than the value of a function at c. If you get an indeterminate form like 0/0 or infinity/infinity, you typically have to do some more math to evaluate the limit. In this case, try factoring, and take into account that the abs function is a piecewise function

Orodruin
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Homework Statement: lim(x->-2) (x^2 +2x) /(1-abs(x+3))
Homework Equations: setting the limit we get 2.

but plugging -2 you clearly get a 0/0 answer. which one is correct?
is x=-2 an asymptote?
The point with limits of the form 0/0 is that you cannot just ”plug in” the value of the variable.

The point with limits of the form 0/0 is that you cannot just ”plug in” the value of the variable.
ok well what happens with the function at EXACTLY f(-2) ? limits we are tending to a point. I want to to know what happens at that exact point. surely its undefined.

PeroK
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ok well what happens with the function at EXACTLY f(-2) ? limits we are tending to a point. I want to to know what happens at that exact point. surely its undefined.
The function is undefined at that point. The limit depends only on values of the function at other points.

The function is undefined at that point. The limit depends only on values of the function at other points.
ok if its undefined at that point then surely x=-2 is an asymptote.

PeroK
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ok if its undefined at that point then surely x=-2 is an asymptote.
Look up the definition of asymptote. That's something different.

$x = -2$ is a vertical line, which is not an asymptote of your function.

Look up the definition of asymptote. That's something different.

$x = -2$ is a vertical line.
which the function cannot cross since its undefined at that point, ergo an asymptote. we can have horizontal, vertical and also y=kx+m asymptotes.

PeroK
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which the function cannot cross since its undefined at that point, ergo an asymptote. we can have horizontal, vertical and also y=kx+m asymptotes.
That's not an asymptote. Look up the definition.

If you've come to learn, believe me that's not an asymptote.

For this function, see if you can find the real asymptotes. There are two.

That's not an asymptote. Look up the definition.

If you've come to learn, believe me that's not an asymptote.

For this function, see if you can find the real asymptotes. There are two.
ya forgot lim fx/x thanks

PeroK
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which the function cannot cross since its undefined at that point, ergo an asymptote. we can have horizontal, vertical and also y=kx+m asymptotes.
What you have is a removable discontinuity at $x = -2$.

What you have is a removable discontinuity at $x = -2$.
can i say the function is undefined at x=-2?

PeroK
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can i say the function is undefined at x=-2?
Yes.

Yes.
so how is that different from using asymptotes when figuring out where the function is defined. so the function is NOT defined at X=-2?

so its defined from (inf, -4) then from (-4,-2) then from (-2, inf) ?

vela
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"asymptote" and "undefined" are not synonyms. Did you do as @PeroK suggested several times and look up the definition of "asymptote"?

WWGD
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If x=-2 were an asymptote, the values of f would become indefinitely close to it. Notice they do not do so in actuality.

so writing the definition parameters x=-2 is undefined? well that is what i've been saying the entire time. if you read the thread im getting mixed signals here.

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Mark44
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ok well what happens with the function at EXACTLY f(-2) ? limits we are tending to a point. I want to to know what happens at that exact point. surely its undefined.
Yes, it's undefined at x = -2.
ok if its undefined at that point then surely x=-2 is an asymptote.
can i say the function is undefined at x=-2?
It's fairly obvious that the function is undefined at x = -2. The whole point of evaluating this limit is to determine whether there is a removable discontinuity at x = -2 (a "hole" in the graph) or that the function's values become unbounded (go to $\infty$ or $-\infty$ -- a vertical asymptote).
so writing the definition parameters x=-2 is undefined? well that is what i've been saying the entire time. if you read the thread im getting mixed signals here.
The signals being sent to you are very clear.

benorin
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Whoa you guys should get hazard pay for this thread! Kudos for keeping a cool heads.

why is y=x-2 an asymptote

but y=-x is not?

the math checks out.

Mark44
Mentor
why is y=x-2 an asymptote
It's not. Why do you think it is?
but y=-x is not?
I agree that y = -x is an oblique asymptote. Does your textbook say otherwise?
the math checks out.

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It's not. Why do you think it is?
I agree that y = -x is an oblique asymptote. Does your textbook say otherwise?
it says y=x-2 is and y=-x is not. is it possible y=x-2 is an asymptote for x<-3 ? and vice versa?

Mark44
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it says y=x-2 is and y=-x is not. is it possible y=x-2 is an asymptote for x<-3 ? and vice versa?
For very large x, I get an oblique asymptote of y = -x. For very negative x, I get an oblique asymptote of y = x - 2, which agrees with what you have.

If x > -3, |x + 3| = x + 3.
If x > -2 (to omit the discontinuity at x = -2),
$$\frac{x^2 + 2x}{1 - |x + 3|} = \frac{x^2 + 2x}{1 - x - 3} = \frac{x^2 + 2x}{- x - 2}$$
Simply by factoring the last expression we can see that if x > -2, we get y = -x.