Paul's Online Notes
Home / Calculus I / Limits / The Definition of the Limit
Show Mobile Notice Show All Notes Hide All Notes
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

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

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

Show All Steps Hide All Steps

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,

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