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### Section 2.10 : The Definition of the Limit

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

$\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty$

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Start Solution

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

Let $$M > 0$$ be any number. We need to find a number $$\delta > 0$$ so that,

$\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 1} \right| < \delta$ Show Step 2

Let’s do a little rewrite the first inequality above a little bit.

$\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.25in}\,\, \to \hspace{0.25in}\,\,\,\,\,{\left( {x - 1} \right)^2} < \frac{1}{M}\hspace{0.25in}\,\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\left| {x - 1} \right| < \frac{1}{{\sqrt M }}$

From this it looks like we can choose $$\delta = \frac{1}{{\sqrt M }}$$.

Show Step 3

So, let’s see if this works.

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

$0 < \left| {x - 1} \right| < \delta = \frac{1}{{\sqrt M }}\hspace{0.5in} \Rightarrow \hspace{0.5in}0 < \left| {x - 1} \right| < \frac{1}{{\sqrt M }}$

\begin{align*}\left| {x - 1} \right| & < \frac{1}{{\sqrt M }} & & \\ & {\left| {x - 1} \right|^2} < \frac{1}{M} & & \hspace{0.25in}{\mbox{squaring both sides}}\\ & {\left( {x - 1} \right)^2} < \frac{1}{M} & & \hspace{0.25in}\,\,{\mbox{because }}{\left| {x - 1} \right|^2} = {\left( {x - 1} \right)^2}\\ & \frac{1}{{{{\left( {x - 1} \right)}^2}}} > M & & \hspace{0.25in}{\mbox{ rewriting things a little bit}}\end{align*}
$\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 1} \right| < \frac{1}{{\sqrt M }}$
$\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty$