# Inequalities/limits question

1. Nov 13, 2008

1. The problem statement, all variables and given/known data

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 $$\frac{1}{x} = 0$$
$$x \rightarrow -\infty$$

I get stuck at the inequality point.

So, let $$\epsilon$$>0, N<0 (N is negative) such that

$$\left|\frac{1}{x} - 0\right| < \epsilon$$ whenever x<N

So, $$x > \frac{1}{\epsilon}$$

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

$$x > -\frac{1}{\epsilon}$$

Now this is where I get confused.

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

Thanks

2. Nov 13, 2008

### HallsofIvy

Staff Emeritus
No. You can't just "add a negative sign". Since we know that know that x is negative, |1/x|= -1/x and $|1/x|< \epsilon$ is immediately $-1/x< \epsilon$ and, multiplying both sides by the negative number $x/\epsilon$, $-1/\epsilon> x$

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.