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

[roadworx]

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 \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| &lt; \epsilon whenever x<N

So, x &gt; \frac{1}{\epsilon}

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

x &gt; -\frac{1}{\epsilon}

Now this is where I get confused.

I have that x &gt; -\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

 

[HallsofIvy]

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 \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| &lt; \epsilon whenever x<N

So, x &gt; \frac{1}{\epsilon}

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

x &gt; -\frac{1}{\epsilon}

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|&lt; \epsilon is immediately -1/x&lt; \epsilon and, multiplying both sides by the negative number x/\epsilon, -1/\epsilon&gt; x

Now this is where I get confused.

I have that x &gt; -\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

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.