roadworx
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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