Exploring the Validity of x=-2 as an Asymptote

[Wi_N]

Homework Statement

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

Relevant 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?

 

[WWGD]

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

 

[PeroK]

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 $x = -2$.

 

[arochon]

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]

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.

 

[Wi_N]

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]

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.

 

[Wi_N]

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]

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.

 

[Wi_N]

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]

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.

 

[Wi_N]

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]

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 $x = -2$.

 

[Wi_N]

What you have is a removable discontinuity at $x = -2$.

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

 

[PeroK]

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

Yes.

 

[Wi_N]

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]

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

 

[WWGD]

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

 

[Wi_N]

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 I am getting mixed signals here.

 

[Mark44]

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 I am getting mixed signals here.

The signals being sent to you are very clear.

 

[benorin]

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

 

[Wi_N]

why is y=x-2 an asymptote

but y=-x is not?

the math checks out.

 

[Mark44]

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.

 

[Wi_N]

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]

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.

 

[Wi_N]

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.

so we have 3 asymptotes.
x=-4 for x<-3 (obvi)
y=x-2 for x<-3
y=-x for x>-3

 

[Mark44]

so we have 3 asymptotes.
x=-4 for x<-3 (obvi)

Just say x = -4 is a vertical asymptote. Don't add "x < -3".

y=x-2 for x<-3
y=-x for x>-3

The latter two are oblique asymptotes. Horizontal and oblique asymptotes have to do with the graph's behavior for very large or very negative x values.

 

[Wi_N]

Just say x = -4 is a vertical asymptote. Don't add "x < -3".
The latter two are oblique asymptotes. Horizontal and oblique asymptotes have to do with the graph's behavior for very large or very negative x values.

thank you.

 

[SammyS]

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.

As you show:

For $x>-2$ the function $\dfrac{x^2 + 2x}{1 - |x+ 3|}$ is identically equal to $x$.

So, for $x>-2$ the graph of our function lies directly on the graph $y=x$. I wouldn't call that asymptotic behavior.

 

[epenguin]

Not writing a waterproof complete explanation which would look more complicated than it is, but roughly on the graph of a function that contains abs if it has a zero and the same function without the abs is continuous, then the function will be continuous but have a kink in it, see blue-green curve in the attachment* at x = - 3.Then the abs incorporating into other functions can give you kinks at points which are not zeros, see dark red curve, which is our denominator.When a denominator is a zero and the numerator is also continuous and non-zero at that same point you have an infinite discontinuity, as you should already know, and as this illustrated in the jade curve at x = -4 which is our function, however if both numerator and denominator have a zero at the same point then usually the quotient will be finite there and continuous but kinked as here at x = -2.

These points are illustrated in the diagram, as also others discussed above, such as the asymptote and the special feature of this example, that for x > -3 , numerator and denominator have a common factor and the equation reduces to y = -x

*The equation for corresponding to each colour of curve is given in the last items in the list on the left, but take no notice of the preceding equations which are from other work that I did not manage to block out.