One sided limits

Although sometimes we get results directly, it is usually helpful to have a larger range of techniques. We give one here and more in section 4.4.

Definition 4.9   Say that $\lim_{{x\to a-}}^{}$f (x) = l, or that f has a limit from the left iff given $\epsilon$ > 0, there is some $\delta$ > 0 such that whenever a - $\delta$ < x < a, then | f (x) - f (a)| < $\epsilon$.

There is a similar definition of ``limit from the right'', writen as $\lim_{{x\to a+}}^{}$f (x) = l

Example 4.10   Define f (x) as follows:-

f (x) = $$\left\{\vphantom{ \begin{array}{ll} 3-x&\mbox{if $x <2$;}\\ 2& \mbox{if $x = 2$;}\\ x/2& \mbox{if $x > 2$.} \end{array}}\right.$$$$\begin{array}{ll} 3-x&\mbox{if $x <2$;}\\ 2& \mbox{if $x = 2$;}\\ x/2& \mbox{if $x > 2$.} \end{array}$$

Calculate the left and right hand limits of f (x) at 2.

Solution. As x$\to$2 -, f (x) = 3 - x$\to$1 +, so the left hand limit is 1. As x$\to$2 +, f (x) = x/2$\to$1 +, so the right hand limit is 1. Thus the left and right hand limits agree (and disagree with f (2), so f is not continuous at 2).

Note our convention: if f (x)$\to$1 and always f (x)$\ge$1 as x$\to$2 -, we say that f (x) tends to 1 from above, and write f (x)$\to$1 + etc.

Proposition 4.11   If $\lim_{{x\to a}}^{}$f (x) exists, then both one sided limts exist and are equal. Conversely, if both one sided limits exits and are equal, then $\lim_{{x\to a}}^{}$f (x) exists.

This splits the problem of finding whether a limit exists into two parts; we can look on either side, and check first that we have a limit, and second, that we get the same answer. For example, in 4.2, example 5, both 1-sided limits exist, but are not equal. There is now an obvious way of checking continuity.

Proposition 4.12   (Continuity Test) The function f is continuous at a iff both one sided limits exits and are equal to f (a).

Example 4.13   Let f (x) = $$\left\{\vphantom{ \begin{array}{ll} x^2& \mbox{for $x \le 1$,}\\ x& \mbox{for $x \ge 1$.} \end{array}}\right.$$$$\begin{array}{ll} x^2& \mbox{for $x \le 1$,}\\ x& \mbox{for $x \ge 1$.} \end{array}$$ Show that f is continuous at 1. [In fact f is continuous everywhere].

Solution. We use the above criterion. Note that f (1) = 1. Also

$$\lim_{{x\to 1-}}^{}$$f (x) = $$\lim_{{x\to 1-}}^{}$$x² = 1    while    $$\lim_{{x\to 1+}}^{}$$f (x) = $$\lim_{{x\to 1+}}^{}$$x = 1 = f (1).

so f is continuous at 1.

Exercise 4.14   Let f (x) = $$\left\{\vphantom{ \begin{array}{ll} \displaystyle \frac{\sin x}{x}& \mbox{for $x < 0$,}\\ \cos x& \mbox{for $x \ge 0$.} \end{array}}\right.$$$$\begin{array}{ll} \displaystyle \frac{\sin x}{x}& \mbox{for $x < 0$,}\\ \cos x& \mbox{for $x \ge 0$.} \end{array}$$ Show that f is continuous at 0. [In fact f is continuous everywhere]. [Recall the result of 4.2, example 4]

Example 4.15   Let f (x) = | x|. Then f is continuous in $\mathbb {R}$.

Solution. Note that if x < 0 then | x| = - x and so is continuous, while if x > 0, then | x| = x and so also is continuous. It remains to examine the function at 0. From these identifications, we see that $\lim_{{x \to 0-}}^{}$| x| = 0 +, while $\lim_{{x \to 0+}}^{}$| x| = 0 +. Since 0 + = 0 - = 0 = | 0|, by the 4.12, | x| is continuous at 0