Let’s do a little rewrite on the first inequality above to get,

$$\frac{1}{x} < N\hspace{0.25in}\,\, \to \hspace{0.25in}\,\,\,\,\,x > \frac{1}{N}$$

Now, keep in mind that $N$ is negative and so ${\textstyle{1 \over N}}$ is also negative. From this it looks like we can choose $\delta = - \frac{1}{N}$. Again, because $N$ is negative this makes $\delta$ positive, which we need!

Show Step 3

So, let’s see if this works.

We’ll start by assuming that $N < 0$ is any number and chose $\delta = - \frac{1}{N}$. We can now assume that,

$$- \delta < x - 0 < 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\frac{1}{N} < x < 0$$

So, if we start with the second inequality we get,

$$\begin{align*}x & > \frac{1}{N} & & \\ & \frac{1}{x} < N & & \hspace{0.25in}{\mbox{ rewriting things a little bit}}\end{align*}$$

So, we’ve shown that,

$$\frac{1}{x} < N\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}\frac{1}{N} < x < 0$$

and so by the definition of the limit we have just proved that,

$$\mathop {\lim }\limits_{x \to {0^ - }} \frac{1}{x} = - \infty$$