What is 0 multiplied by infinity in limits?

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Discussion Overview

The discussion revolves around the concept of multiplying limits that approach 0 and infinity, particularly in the context of indeterminate forms in calculus. Participants explore whether such multiplications yield meaningful results or if they require further evaluation, focusing on specific functions and their limits as they approach certain values.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that multiplying limits that approach 0 and infinity leads to an indeterminate form, indicating that further evaluation is necessary.
  • Others argue that the expression formed by multiplying two limits cannot be split into separate limits meaningfully, emphasizing the need to evaluate the limit of the product directly.
  • A participant points out that the limit of f(x) = 1/x as x approaches 0 is infinity, and highlights the importance of one-sided limits, noting that they do not agree in this case.
  • Another participant introduces a new example involving different variables, questioning if the limits can be evaluated despite one approaching infinity and the other approaching 0.
  • Some responses reiterate that the presence of different variables does not resolve the indeterminate nature of the limits.

Areas of Agreement / Disagreement

Participants express disagreement regarding the treatment of limits approaching 0 and infinity, with some asserting that the multiplication is indeterminate while others emphasize the necessity of evaluating the product directly. The discussion remains unresolved, with multiple competing views on how to approach the problem.

Contextual Notes

Limitations include the dependence on definitions of limits and the unresolved nature of one-sided limits in the context of the two-sided limit existing or not.

MathewsMD
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If you have f(x) = 1/x and g(a) = (cosx - 1)/x and then y = [limx→0 f(x)][limx→0g(x)], the two individual limits equal 0 and infinity, respectively. Since these are limits and only approach these values, would the multiplication of the two limits equal 0, infinity or something else?
 
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In this case, y= \lim_{x\to 0} f(x) \lim_{x\to 0} g(x) is a meaningless expression. You CAN calculate
\lim_{x\to 0} f(x) g(x)
but you cannot split it into two limits and give the new expression any meaning.
 
If you take a limit and get 0 * ∞, it means that you're not done yet. The indeterminate form [0 * ∞] is one of several indeterminate forms that can arise when you're taking limits. One thing that is often done is to rewrite the expression so that the limit becomes [0/0] or [∞/∞], either of which might be amenable to evaluation using L'Hopital's Rule.

This wiki article has a table of indeterminate forms - http://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
 
MathewsMD said:
If you have f(x) = 1/x and g(a) = (cosx - 1)/x and then y = [limx→0 f(x)][limx→0g(x)], the two individual limits equal 0 and infinity, respectively. Since these are limits and only approach these values, would the multiplication of the two limits equal 0, infinity or something else?

The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.
 
SteamKing said:
The limit of f(x) = 1/x as x approaches 0 is not zero, it is infinity.
You're sort of half correct.
$$\lim_{x \to 0^+}\frac 1 x = ∞$$
$$\lim_{x \to 0^-}\frac 1 x = -∞$$
Since the one-sided limits aren't equal, the two-sided limit doesn't exist.
 
Sorry for my poor phrasing.

How about in this new example.

y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.
 
Last edited:
MathewsMD said:
Sorry for my poor phrasing.

How about in this new example.

y = [limx→0^+ 1/x][limz→0 (1 - cosz)/z]

Would it be possible to evaluate this limit since one limit approaches infinity while the other approaches 0, and they are different variables in this case.
No, you're right back to what I was talking about in post #3, and what you had in your first post. The fact that you have different variables does not change things.
 
How did this limit come up? What's the problem you're trying to solve?
 

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