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Section 2.10 : The Definition of the Limit

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

\[\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\]

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Start Solution

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

Let \(\varepsilon > 0\) be any number. We need to find a number \(M > 0\) so that,

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

Or, with a little simplification this becomes,

\[\left| {\frac{1}{{{x^2}}}} \right| < \varepsilon \hspace{0.5in}{\mbox{whenever}}\hspace{0.5in}x > M\] Show Step 2

Let’s start with the inequality on the left and do a little rewriting on it.

\[\left| {\frac{1}{{{x^2}}}} \right| < \varepsilon \hspace{0.25in} \to \hspace{0.25in}\frac{1}{{{{\left| x \right|}^2}}} < \varepsilon \hspace{0.25in}\, \to \hspace{0.25in}{\left| x \right|^2} > \frac{1}{\varepsilon }\hspace{0.25in} \to \hspace{0.25in}\,\left| x \right| > \frac{1}{{\sqrt \varepsilon }}\]

From this it looks like we can choose \(M = \frac{1}{{\sqrt \varepsilon }}\)

Show Step 3

So, let’s see if this works.

Start off by first assuming that \(\varepsilon > 0\) is any number and choose \(M = \frac{1}{{\sqrt \varepsilon }}\). We can now assume that,

\[x > \frac{1}{{\sqrt \varepsilon }}\]

Starting with this inequality we get,

\[\begin{align*}x & > \frac{1}{{\sqrt \varepsilon }} & & \\ & \frac{1}{x} < \sqrt \varepsilon & & \hspace{0.25in}{\mbox{do a little rewrite}}\\ & \frac{1}{{{x^2}}} < \varepsilon & & \hspace{0.25in}{\mbox{square both sides}}\\ & \left| {\frac{1}{{{x^2}}}} \right| < \varepsilon & & \hspace{0.25in}{\mbox{because }}\frac{1}{{{x^2}}} = \left| {\frac{1}{{{x^2}}}} \right|\end{align*}\]

So, we’ve shown that,

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

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

\[\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{x^2}}} = 0\]