Proving a limit is false when L does not equal 1

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Homework Help Overview

The problem involves proving that if \( L \neq 1 \), the limit statement \( \lim \limits_{x \to \infty} (1+\frac{1}{x}) = L \) is false. The context is centered around the definition of limits in calculus.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to negate the definition of a limit to show the statement is false, but struggles to find a suitable \( x \) for their proof. Some participants suggest using limit laws and manipulating the absolute value to explore the behavior of the function as \( x \) approaches infinity.

Discussion Status

Participants are actively discussing various approaches to the problem, including suggestions for choosing specific values for \( \epsilon \) and manipulating inequalities. There is a focus on exploring different cases for \( L \) and how to effectively apply the triangle inequality. No consensus has been reached yet, and the discussion remains open-ended.

Contextual Notes

There is an emphasis on the need to consider different cases for \( L \) (greater than or less than 1) and the challenges associated with manipulating absolute values in the context of limits. The original poster expresses uncertainty about their algebraic manipulations.

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Homework Statement


Show that if [itex]L \neq 1[/itex], the statement [tex]\lim \limits_{x \to \infty} (1+\frac{1}{x}) = L[/tex] is false.


Homework Equations



The Definition of a Limit

The Attempt at a Solution


So I've been trying to prove this by negating the logical statement of the definition of a limit; i.e. by trying to prove that
[itex]\exists \epsilon > 0[/itex] such that [itex]\forall \delta >0 \exists x > \delta[/itex] such that [itex]\left|f(x)-L\right|\geq \epsilon[/itex].
I know that when [itex]L=1[/itex] the limit exists; that is no trouble to prove. The problem is that every time i try to find an x that works, I can never make it work in my proof. Am I going about this the right way?
 
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Use the limit Law to break up your limit
 
Think about it this way. Suppose L > 1. Now you know how to get the function very close to 1, so you should be able to keep it away from L, no?. So think about what would happen if you choose epsilon half the distance from L to 1. And the case L < 1 is similar or you could combine them.
 
So if I choose [itex]\epsilon = \frac{L-1}{2}[/itex], I want to find an x that will give me [tex]\left|f(x)-L\right|=\left|1+\frac{1}{x}-L\right|=\epsilon=\frac{L-1}{2}.[/tex]

This is kind of where I'm stuck. I'm not sure what the best way to manipulate the absolute value sign. I know I can do it with either the triangle or reverse triangle inequality, but which would be the right direction to take? would it matter?
 
Use |L-1|/2, in case L < 1.
 
ok, so starting with [tex]\left|1+\frac{1}{x}-L\right|\geq\frac{\left| L-1 \right|}{2},[/tex]
I get
[tex]\left| 1 \right| + \left| \frac{1}{x} - L \right| \geq \frac{\left| L-1 \right|}{2} \text{(triangle inequality)}[/tex]

[tex]\left| \frac{1}{x} - L \right| \geq \frac{\left| L-1 \right|}{2} -1 = \frac{\left| L-1 \right|-\left|2\right|}{2} \geq \frac{\left| L-3 \right|}{2} \text{(reverse triangle ineq.)}[/tex]

So now I'm stuck here at
[tex]\left| \frac{1}{x} - L \right| \geq \frac{\left| L-3 \right|}{2} .[/tex]

I'm not too experienced with doing algebra with absolute value signs. Have I been doing it right so far?
 

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