How Can You Find Functions with Indeterminate Limits at Infinity?

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Homework Help Overview

The discussion revolves around finding functions f and g that both approach zero as x approaches infinity, while the limit of their ratio approaches positive infinity. This is a topic within the subject area of calculus, specifically dealing with limits and indeterminate forms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's attempts to find suitable functions and question the validity of their choices, particularly regarding the behavior of g(x). There is also exploration of how to systematically approach the problem of finding functions that meet the specified criteria.

Discussion Status

Some participants have provided guidance on the nature of the functions needed, suggesting that one function should approach zero more slowly than the other. The conversation reflects a mix of attempts and clarifications, with no explicit consensus reached on a specific solution.

Contextual Notes

Participants note that the original functions proposed do not satisfy the condition of approaching zero, and there is an acknowledgment of the challenge in finding appropriate functions that meet the problem's requirements.

LemuelUhuru
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Hi, I am in a first semester Calculus I course in college with an intermediate skill level with precalc and a basic understanding of limits and infinity. I do not understand how to solve this problem I attempted to do so only to find out after completion that ∞/∞ is indeterminate rendering my solution void.

Homework Statement



Suppose f(x) →0 and g(x) → 0 as x → +∞. Find examples of functions f and g with these properties and such that:
a. limx→+∞[itex]\left[\frac{f(x)}{g(x)}\right][/itex] = + ∞

The Attempt at a Solution



Here is my attempt

Let f(x) = 1/x and g(x) = (1/x)+1
limx→+∞ [itex]\left[\frac{\frac{1}{x}}{\frac{1}{x}+1}\right][/itex]= [itex]\frac{1}{x}[/itex] * [itex]\frac{x}{1}+1[/itex] = ∞ + 1 = + ∞
 
Last edited:
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LemuelUhuru said:
Hi, I am in a first semester Calculus I course in college with an intermediate skill level with precalc and a basic understanding of limits and infinity. I do not understand how to solve this problem I attempted to do so only to find out after completion that ∞/∞ is indeterminate rendering my solution void.

Homework Statement



Suppose f(x) →0 and g(x) → 0 as x → +∞. Find examples of functions f and g with these properties and such that:
a. limx→+∞[itex]\left[\frac{f(x)}{g(x)}\right][/itex] = + ∞

The Attempt at a Solution



Here is my attempt

Let f(x) = 1/x and g(x) = (1/x)+1
First problem, g(x) doesn't go to 0 in the limit as ##x \to \infty##.

limx→+∞[itex]\left[\frac{\frac{1}{x}}{\frac{1}{\frac{1}{x}+1}}\right][/itex] = [itex]\frac{1}{x}[/itex] * [itex]\frac{x}{1}+1[/itex] = ∞ + 1 = + ∞
I'm not sure what you did here because you have f(x) in the numerator but 1/g(x) in the denominator, and then you did some funky "algebra."
 
Hi Vela,

I realize now that g(x) doesn't go to zero, that was a huge error and I adjusted the original attempt with respect to 1/g(x), although that despite it being irrelevant now, however, I am still unsure how to approach this problem generally. I imagine that I am suppose to obtain to two functions both f(x) and g(x) who limit approaches zero and when divided result in a/0 to give me positive infinity?

Thank you,

Lemuel
 
Yup, that's the idea.
 
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I understand, however, right now I'm simply guessing what rational expressions would allow me to do this, is there a more systematic approach to solving this kind of problem?

Also, it seems difficult to obtain a/0 since the limit of f(x) and g(x) as x approaches infinity is zero and any two functions where a/∞ cancels out my constant when x is subtituted for infinity.
 
Last edited:
All you're looking for is for g(x) to be something that goes to zero really fast, and f(x) to be something that goes to zero slower.

If you're not sure how to do that, make f(x) = 1/F(x) and g(x) = 1/G(x) where F(x) and G(x) go to infinity, but G(x) much faster.
 

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