Proving a limit is false when L does not equal 1

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SUMMARY

The discussion centers on proving that if L ≠ 1, then the limit statement \(\lim_{x \to \infty} (1+\frac{1}{x}) = L\) is false. The user attempts to negate the definition of a limit by showing the existence of an epsilon (\(\epsilon\)) such that for all delta (\(\delta\)), there exists an x that satisfies \(\left|f(x)-L\right|\geq \epsilon\). The discussion highlights the use of the triangle and reverse triangle inequalities to manipulate absolute values in the proof. The user struggles with finding the correct x to demonstrate the limit's falsity when L is not equal to 1.

PREREQUISITES
  • Understanding of limit definitions in calculus
  • Familiarity with epsilon-delta proofs
  • Knowledge of triangle and reverse triangle inequalities
  • Basic algebraic manipulation of absolute values
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  • Study the epsilon-delta definition of limits in detail
  • Practice problems involving the triangle inequality in calculus
  • Explore examples of limits that do not equal 1
  • Learn advanced techniques for manipulating absolute values in proofs
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Students studying calculus, particularly those focusing on limits and epsilon-delta proofs, as well as educators looking for examples of limit negation techniques.

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Homework Statement


Show that if L \neq 1, the statement \lim \limits_{x \to \infty} (1+\frac{1}{x}) = L 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
\exists \epsilon > 0 such that \forall \delta >0 \exists x > \delta such that \left|f(x)-L\right|\geq \epsilon.
I know that when L=1 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 \epsilon = \frac{L-1}{2}, I want to find an x that will give me \left|f(x)-L\right|=\left|1+\frac{1}{x}-L\right|=\epsilon=\frac{L-1}{2}.

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 \left|1+\frac{1}{x}-L\right|\geq\frac{\left| L-1 \right|}{2},
I get
\left| 1 \right| + \left| \frac{1}{x} - L \right| \geq \frac{\left| L-1 \right|}{2} \text{(triangle inequality)}

\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.)}

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

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

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