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,

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

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

\[\begin{align*}\left| {x - 1} \right| & < \frac{1}{{\sqrt M }} & & \\ & {\left| {x - 1} \right|^2} < \frac{1}{M} & & \hspace{0.25in}{\mbox{squaring both sides}}\\ & {\left( {x - 1} \right)^2} < \frac{1}{M} & & \hspace{0.25in}\,\,{\mbox{because }}{\left| {x - 1} \right|^2} = {\left( {x - 1} \right)^2}\\ & \frac{1}{{{{\left( {x - 1} \right)}^2}}} > M & & \hspace{0.25in}{\mbox{ rewriting things a little bit}}\end{align*}\]

So, we’ve shown that,

\[\frac{1}{{{{\left( {x - 1} \right)}^2}}} > M\hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}0 < \left| {x - 1} \right| < \frac{1}{{\sqrt M }}\]

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

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