Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 2.10 : The Definition of the Limit
5. Use the definition of the limit to prove the following limit.
\[\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty \]Show All Steps Hide All Steps
Start SolutionFirst, 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 2Let’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 3So, 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,
\[\mathop {\lim }\limits_{x \to 1} \frac{1}{{{{\left( {x - 1} \right)}^2}}} = \infty \]