Limit approaching infinity with absolute function

  • Thread starter physicoo
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  • #1
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Hi guys,

I have a little problem with this question:

Determine the limit (x approaching infinity) [tex]\frac{|9-3x^{2}|}{x-2x^{2}}[/tex]

(Sorry, I'm not sure on how to use the Latex feature)

Anyways, I've combed the forum and found a couple of threads that are somewhat related to my problem.

Yes, I've found the two one-sided limits, and I got -3/2 and 3/2. I know for one, that if they aren't the same, the limit for a function does not exist. However, does this apply to this case, where x approaches infinity?

Would appreciate if someone could guide me on this :)

Cheers!
 

Answers and Replies

  • #2
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Not sure how to find the limit, but how can it be 3/2? abs() is always positive and (x-2x^2) is negative for large x.
 
  • #3
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it's through using one of the limit methods, that is to divide both the numerator and dominator by the highest power which in this case is x^2
 
  • #4
uart
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it's through using one of the limit methods, that is to divide both the numerator and dominator by the highest power which in this case is x^2
You should remove the abs value at the first step. Clearly [itex]9-3x^2[/itex] is negative as x goes to infinity, therefore the absolute value can be replaced with the negative of it's argument.
 
  • #5
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May I know what's the purpose of removing the modulus? I'm not quite sure what you mean by

You should remove the abs value at the first step. Clearly [itex]9-3x^2[/itex] is negative as x goes to infinity, therefore the absolute value can be replaced with the negative of it's argument.
Sorry. It's a little confusing to me. I was referring to this thread:

https://www.physicsforums.com/showthread.php?t=115063

while solving my problem. But I wasn't sure if it was the same concept when it comes to x approaching infinity. In that thread, it was x approaching 1.
 
Last edited:
  • #6
uart
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May I know what's the purpose of removing the modulus? I'm not quite sure what you mean by
[tex] |u| = +u : u \ge 0[/tex]

[tex] |u| = -u : u<0[/tex]

So if you know that [itex]u[/itex] is less than zero (in the region of interest) then just drop the absolute value and replace it with [itex]-u[/itex].
 
  • #7
Hurkyl
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The limit as [itex]x \to +\infty[/itex] is a left-sided limit -- there is no such thing as a right sided limit at [itex]+\infty[/itex].

I expect you didn't compute one-sided limits -- I expect what you actually did is:

  • You separated the real line into two regions: one where [itex]x^2 \leq 3[/itex] and one where [itex]x^2 \geq 3[/itex]
  • On the first region, you simplified the expression into a rational function, and you took the limit of that rational function as [itex]x \to +\infty[/itex]
  • On the second region, you simplified the expression into a rational function, and you took the limit of that rational function as [itex]x \to +\infty[/itex]

If you think about what you're doing, one of these calculations obviously gives the limit as [itex]x \to +\infty[/itex], and the other calculation is obviously irrelevant....
 
  • #8
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The limit as [itex]x \to +\infty[/itex] is a left-sided limit -- there is no such thing as a right sided limit at [itex]+\infty[/itex].

I expect you didn't compute one-sided limits -- I expect what you actually did is:

  • You separate the real line into two regions: one where [itex]x^2 \leq 3[/itex] and one where [itex]x^2 \geq 3[/itex]
  • On the first region, you simplified the expression into a rational function, and you took the limit of that rational function as [itex]x \to +\infty[/itex]
  • On the second region, you simplified the expression into a rational function, and you took the limit of that rational function as [itex]x \to +\infty[/itex]

If you think about what you're doing, one of these calculations obviously gives the limit as [itex]x \to +\infty[/itex], and the other calculation is obviously irrelevant....
You're right in a way!

Yes I actually did separated them into two to find both side limits. Like I said, I was referring to the thread that I had pasted earlier on. But I was unsure if it was correct since, like you said, when x approaches infinity, it is a left-sided limit, which is the reason why I had doubts with my own answer.

I suppose the one approaching from the left is the relevant one and the other is not. However, if this is the case, does the limit still exist? Because to my knowledge, a limit only exists when both side limits are equal. In this case, one is irrelevant.

Cheers!
 
  • #9
Hurkyl
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I wonder if you have a terminology mix-up?

The right-sided limit at a is called such because its epsilon-delta definition is
For all [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that for every x with [itex]a < x < a + \delta[/itex], we have [itex]|f(x) - L| < \epsilon[/itex]​
In other words, x is restricted to being on the right of a on the number line.



I get the feeling you are thinking of the sidedness as relating to an absolute value appearing in the expression -- e.g. thinking of the two sides as being where [itex]9 - 3x^2[/itex] is positive and where [itex]9 - 3x^2[/itex] is negative.

If you were computing the limit as [itex]x \to \sqrt{3}[/itex] (or the limit as [itex]x \to -\sqrt{3}[/itex]), then the two are closely related. For [itex]\delta[/itex] chosen sufficiently small, the ordinary limit looks involves the set of x with [itex]0 < |x - \sqrt{3}| < \delta[/itex]. The left and right-sided limits break this region up into whether or not x is less than or greater than [itex]\sqrt{3}[/itex] respectively, and these happen to be exactly the regions where [itex]9 - 3x^2[/itex] is positive or negative, respectively.


However, in the problem you're actually solving, there is no such relationship between the point x is approaching and the argument of the absolute value.



Because to my knowledge, a limit only exists when both side limits are equal.
Since this piece of knowledge is involved in your conflict, it would be worth checking up on the theorem that says such. In this case, there's a hypothesis you're forgetting -- the statement you've made only applies to limits as [itex]x \to a[/itex] for a real number a.
 
  • #10
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I guess I have to relook into the question and theorems.

Thanks for the advice so far :)
 

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