Prove [itex]\lim_{a\to 0}\frac{1}{a} = \infty[/itex]

  • Thread starter Thread starter operationsres
  • Start date Start date
Click For Summary
SUMMARY

The limit proof for \(\lim_{a\to 0^+}\frac{1}{a} = +\infty\) requires using the formal \(\epsilon\)-\(\delta\) definition of limits. The key approach involves demonstrating that for any \(M > 0\), there exists a \(\delta > 0\) such that if \(0 < |a - 0| < \delta\), then \(\frac{1}{a} > M\). This establishes that the function \(\frac{1}{a}\) grows without bound as \(a\) approaches zero from the positive side.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with the \(\epsilon\)-\(\delta\) definition of limits
  • Basic algebraic manipulation of inequalities
  • Concept of functions approaching infinity
NEXT STEPS
  • Study the \(\epsilon\)-\(\delta\) definition of limits in detail
  • Practice proving limits involving infinity using similar techniques
  • Explore examples of limits that approach zero and infinity
  • Learn about the behavior of rational functions near their asymptotes
USEFUL FOR

Students studying calculus, particularly those focusing on limits and continuity, as well as educators looking for clear examples of limit proofs involving infinity.

operationsres
Messages
99
Reaction score
0

Homework Statement

Prove [itex]\lim_{a\to 0^+}\frac{1}{a} = +\infty[/itex] under the [itex]\epsilon[/math] definition of a limit.<br /> <br /> <b>2. The attempt at a solution</b><br /> <br /> Well, I can't do [itex]\frac{1}{a} - \infty < \epsilon[/itex] can I? Otherwise it's just obvious that it's infinity ..[/itex]
 
Physics news on Phys.org
For this particular problem you need to alter your definition abit since |f(x) - ∞| < ε translates into a useless statement.

You want to use this definition :

[itex]\forall M>0, \exists δ>0 \space | \space 0<|x-c|<δ \Rightarrow f(x) > M[/itex]

What this definition essentially means is that we can find a delta such that the function grows without bound.

Start by massaging the expression f(x) > M into a suitable form |x-c| < δ which will give you a δ which MIGHT work.

Then take that δ and show that it implies f(x) > M.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K