- #1
Wi_N
- 119
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- 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?
is x=-2 an asymptote?
The limit is of the form ##0/0## which is why it's non trivial to evaluate.Wi_N said: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 said: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?
Orodruin said:The point with limits of the form 0/0 is that you cannot just ”plug in” the value of the variable.
The function is undefined at that point. The limit depends only on values of the function at other points.Wi_N said: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.
ok if its undefined at that point then surely x=-2 is an asymptote.PeroK said:The function is undefined at that point. The limit depends only on values of the function at other points.
Look up the definition of asymptote. That's something different.Wi_N said:ok if its undefined at that point then surely x=-2 is an asymptote.
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 said:Look up the definition of asymptote. That's something different.
##x = -2## is a vertical line.
That's not an asymptote. Look up the definition.Wi_N said: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 said: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.
What you have is a removable discontinuity at ## x = -2##.Wi_N said: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 said:What you have is a removable discontinuity at ## x = -2##.
Yes.Wi_N said:can i say the function is undefined at x=-2?
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?PeroK said:Yes.
Yes, it's undefined at x = -2.Wi_N said: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.
Wi_N said:ok if its undefined at that point then surely x=-2 is an asymptote.
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).Wi_N said:can i say the function is undefined at x=-2?
The signals being sent to you are very clear.Wi_N said: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.
It's not. Why do you think it is?Wi_N said:why is y=x-2 an asymptote
I agree that y = -x is an oblique asymptote. Does your textbook say otherwise?Wi_N said:but y=-x is not?
Wi_N said:the math checks out.
Mark44 said:It's not. Why do you think it is?
I agree that y = -x is an oblique asymptote. Does your textbook say otherwise?
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.Wi_N said: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?
so we have 3 asymptotes.Mark44 said: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.
Just say x = -4 is a vertical asymptote. Don't add "x < -3".Wi_N said:so we have 3 asymptotes.
x=-4 for x<-3 (obvi)
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 said:y=x-2 for x<-3
y=-x for x>-3
Mark44 said: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.
As you show:Mark44 said: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.
An asymptote is a line that a graph approaches but never touches. It can be horizontal, vertical, or slanted.
To determine if x=-2 is an asymptote, you would need to graph the function and observe the behavior of the graph as x approaches -2. If the graph approaches a specific value as x gets closer to -2, then x=-2 is an asymptote.
If x=-2 is an asymptote, it means that the graph of the function will never touch the line x=-2. This could indicate a discontinuity in the function or a limit that the function approaches as x gets closer to -2.
To prove the validity of x=-2 as an asymptote, you would need to use mathematical techniques such as limits and continuity. You could also use a graphing calculator or software to visually observe the behavior of the graph near x=-2.
Yes, x=-2 can be both a vertical and horizontal asymptote. This would indicate that the graph approaches a specific value as both x and y approach -2. This could happen if the function has a vertical and horizontal asymptote intersecting at the point (-2,-2).