How Do You Prove a Limit Using Epsilon-N Notation for Negative Infinity?

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

The discussion focuses on proving the limit of the function lim (1/x) = 0 as x approaches negative infinity using epsilon-N notation. The key point of confusion arises from the manipulation of inequalities, particularly when defining N as -1/ε. Participants clarify that the correct approach involves recognizing that |1/x| = -1/x for negative x, leading to the inequality -1/ε > x. This ensures that x remains less than N while satisfying the epsilon condition.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with epsilon-N notation
  • Knowledge of absolute value properties
  • Basic algebraic manipulation of inequalities
NEXT STEPS
  • Study the properties of limits approaching negative infinity
  • Learn more about epsilon-delta definitions of limits
  • Explore examples of limits involving negative values
  • Review algebraic manipulation techniques for inequalities
USEFUL FOR

Students in calculus, mathematics educators, and anyone seeking to deepen their understanding of limit proofs using epsilon-N notation.

roadworx
Messages
20
Reaction score
0

Homework Statement



I'm proving the limit of an equation with the epsilon-N notation for negative infinity.

Here is the equation that I'm trying to prove.
lim [tex]\frac{1}{x} = 0[/tex]
[tex]x \rightarrow -\infty[/tex]

I get stuck at the inequality point.

So, let [tex]\epsilon[/tex]>0, N<0 (N is negative) such that

[tex]\left|\frac{1}{x} - 0\right| < \epsilon[/tex] whenever x<N

So, [tex]x > \frac{1}{\epsilon}[/tex]

But we want N to be negative, so add a negative sign

[tex]x > -\frac{1}{\epsilon}[/tex]

Now this is where I get confused.

I have that [tex]x > -\frac{1}{\epsilon}[/tex], yet I want x<N. If N is [tex]-\frac{1}{\epsilon}[/tex], then x not less than N. If x is not less than N, then wouldn't it be wrong to use [tex]-\frac{1}{\epsilon}[/tex] for N, because our x<N has not been satisfied. Or have I got an inequality wrong somewhere?

Thanks
 
Physics news on Phys.org
roadworx said:

Homework Statement



I'm proving the limit of an equation with the epsilon-N notation for negative infinity.

Here is the equation that I'm trying to prove.
lim [tex]\frac{1}{x} = 0[/tex]
[tex]x \rightarrow -\infty[/tex]

I get stuck at the inequality point.

So, let [tex]\epsilon[/tex]>0, N<0 (N is negative) such that

[tex]\left|\frac{1}{x} - 0\right| < \epsilon[/tex] whenever x<N

So, [tex]x > \frac{1}{\epsilon}[/tex]

But we want N to be negative, so add a negative sign

[tex]x > -\frac{1}{\epsilon}[/tex]
No. You can't just "add a negative sign". Since we know that know that x is negative, |1/x|= -1/x and [itex]|1/x|< \epsilon[/itex] is immediately [itex]-1/x< \epsilon[/itex] and, multiplying both sides by the negative number [itex]x/\epsilon[/itex], [itex]-1/\epsilon> x[/itex]

Now this is where I get confused.

I have that [tex]x > -\frac{1}{\epsilon}[/tex], yet I want x<N. If N is [tex]-\frac{1}{\epsilon}[/tex], then x not less than N. If x is not less than N, then wouldn't it be wrong to use [tex]-\frac{1}{\epsilon}[/tex] for N, because our x<N has not been satisfied. Or have I got an inequality wrong somewhere?

Thanks
Yes, you have the inequality wrong. "Add a negative" is not an algebraic operation. You have to be more careful with your manipulation of the absolute value.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
17
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K