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### Section 2.10 : The Definition of the Limit

6. Use the definition of the limit to prove the following limit.

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

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Start Solution

First, let’s just write out what we need to show.

Let $$N < 0$$ be any number. Remember that because our limit is going to negative infinity here we need $$N$$ to be negative. Now, we need to find a number $$\delta > 0$$ so that,

$\frac{1}{x} < N\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in} - \delta < x - 0 < 0$ Show Step 2

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,

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

\begin{align*}x & > \frac{1}{N} & & \\ & \frac{1}{x} < N & & \hspace{0.25in}{\mbox{ rewriting things a little bit}}\end{align*}
$\frac{1}{x} < N\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}\frac{1}{N} < x < 0$
$\mathop {\lim }\limits_{x \to {0^ - }} \frac{1}{x} = - \infty$