2.5 The precise definition of a limit  (Page 5/8)

Showing that a limit does not exist

Show that $\underset{x\to 0}{\text{lim}}\frac{\left|x\right|}{x}$ does not exist. The graph of $f\left(x\right)=\left|x\right|\text{/}x$ is shown here:

Suppose that L is a candidate for a limit. Choose $\epsilon =1\text{/}2.$

Let $\delta >0.$ Either $L\ge 0$ or $L<0.$ If $L\ge 0,$ then let $x=-\delta \text{/}2.$ Thus,

$\left|x-0\right|=\left|-\frac{\delta }{2}-0\right|=\frac{\delta }{2}<\delta$

and

$\left|\frac{\left|-\frac{\delta }{2}\right|}{-\frac{\delta }{2}}-L\right|=\left|\mathrm{-1}-L\right|=L+1\ge 1>\frac{1}{2}=\epsilon .$

On the other hand, if $L<0,$ then let $x=\delta \text{/}2.$ Thus,

$\left|x-0\right|=\left|\frac{\delta }{2}-0\right|=\frac{\delta }{2}<\delta$

and

$\left|\frac{\left|\frac{\delta }{2}\right|}{\frac{\delta }{2}}-L\right|=\left|1-L\right|=\left|L\right|+1\ge 1>\frac{1}{2}=\epsilon .$

Thus, for any value of L , $\underset{x\to 0}{\text{lim}}\frac{\left|x\right|}{x}\ne L.$

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One-sided and infinite limits

Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality $0<x-a<\delta$ replaces $0<\left|x-a\right|<\delta ,$ which ensures that we only consider values of x that are greater than (to the right of) a . Similarly, in the definition of the limit from the left, the inequality $-\delta <x-a<0$ replaces $0<\left|x-a\right|<\delta ,$ which ensures that we only consider values of x that are less than (to the left of) a .

Proving a statement about a limit from the right

Prove that $\underset{x\to 4^{+}}{\text{lim}}\sqrt{x-4}=0.$

Let $\epsilon >0.$

Choose $\delta ={\epsilon }^2.$ Since we ultimately want $\left|\sqrt{x-4}-0\right|<\epsilon ,$ we manipulate this inequality to get $\sqrt{x-4}<\epsilon$ or, equivalently, $0<x-4<{\epsilon }^2,$ making $\delta ={\epsilon }^2$ a clear choice. We may also determine $\delta$ geometrically, as shown in [link] .

Assume $0<x-4<\delta .$ Thus, $0<x-4<{\epsilon }^2.$ Hence, $0<\sqrt{x-4}<\epsilon .$ Finally, $\left|\sqrt{x-4}-0\right|<\epsilon .$

Therefore, $\underset{x\to 4^{+}}{\text{lim}}\sqrt{x-4}=0.$

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We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have $\underset{x\to a}{\text{lim}}f\left(x\right)=\text{+}\infty ,$ we want the values of the function $f\left(x\right)$ to get larger and larger as x approaches a . Instead of the requirement that $\left|f\left(x\right)-L\right|<\epsilon$ for arbitrarily small ε when $0<\left|x-a\right|<\delta$ for small enough $\delta ,$ we want $f\left(x\right)>M$ for arbitrarily large positive M when $0<\left|x-a\right|<\delta$ for small enough $\delta .$ [link] illustrates this idea by showing the value of $\delta$ for successively larger values of M .

Key concepts

• The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-delta definition of the limit .
• The epsilon-delta definition may be used to prove statements about limits.
• The epsilon-delta definition of a limit may be modified to define one-sided limits.