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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,

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

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,

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