Find Limit: Cancellation Homework Statement

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Homework Help Overview

The discussion revolves around finding the limit of the expression \(\lim_{x\rightarrow0}\frac{\frac{1}{x+1}-1}{x}\). Participants are exploring methods to resolve the indeterminate form encountered when substituting \(x = 0\).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to manipulate the expression analytically but encounters a \(0/0\) form. They question how to simplify the function to a determinate form and express uncertainty about the appropriate multiplication to apply. Other participants suggest using a common denominator and multiplying by \(\frac{x+1}{x+1}\) to facilitate cancellation. There is also mention of L'Hopital's rule as a potential approach, although one participant indicates they are unfamiliar with it.

Discussion Status

The discussion is active, with participants providing suggestions for manipulation of the expression. Some guidance has been offered regarding the use of common denominators and the potential application of L'Hopital's rule. There is no explicit consensus on a single method, but multiple avenues are being explored.

Contextual Notes

Participants note the challenge of dealing with the indeterminate form and the limitations of their current calculus knowledge, which may affect their approach to the problem.

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Homework Statement


Find the limit.

[tex]\lim_{x\rightarrow0}\frac{\frac{1}{x+1}-1}{x}[/tex]

Homework Equations


The Attempt at a Solution



I have to do this analytically. Although, I know that the limit is supposed to be -1 from a graphing approach. When you substitute in 0 for x you get 0/0. How do I get this function into a determinate form? I can't factor and I'm not sure what to multiply the top and bottom by.

I tried [tex]\frac{\frac{1}{x+1}+1}{\frac{1}{x+1}+1}[/tex]

and I got

[tex]\frac{1-(x^2+2x+4)}{x(x+1)(x+2)}[/tex] which is still in indeterminate form.
 
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themadhatter1 said:

Homework Statement


Find the limit.

[tex]\lim_{x\rightarrow0}\frac{\frac{1}{x+1}-1}{x}[/tex]

Homework Equations





The Attempt at a Solution



I have to do this analytically. Although, I know that the limit is supposed to be -1 from a graphing approach. When you substitute in 0 for x you get 0/0. How do I get this function into a determinate form? I can't factor and I'm not sure what to multiply the top and bottom by.

I tried [tex]\frac{\frac{1}{x+1}+1}{\frac{1}{x+1}+1}[/tex]
I wouldn't do that. Try multiplying the original fraction by
[tex]\frac{x+1}{x+1}[/tex]
instead. Something should eventually cancel.


69
 
Do you know, or can you use, L'Hopital's rule?

In this case, L'Hopital's rule would take the form

If
[tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{0}{0}[/tex]
but
[tex]\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
where L is a finite number,
then
[tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
 
You can get a common denominator with the terms in the numerator and simplify, then you should get a fraction that's not indeterminate anymore.
 
Mute said:
Do you know, or can you use, L'Hopital's rule?

In this case, L'Hopital's rule would take the form

If
[tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{0}{0}[/tex]
but
[tex]\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]
where L is a finite number,
then
[tex]\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \lim_{x\rightarrow a} \frac{f'(x)}{g'(x)} = L[/tex]

No I dont, I'm just starting calculus, but that looks interesting. I'm sure I'll learn it later.
eumyang said:
I wouldn't do that. Try multiplying the original fraction by
[tex]\frac{x+1}{x+1}[/tex]
instead. Something should eventually cancel.69

Thanks, that works perfectly. I know where I went wrong.
 

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