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

  • Thread starter Thread starter roadworx
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 5K views
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.