One sided limit

When

approaches $x_{0}$ with the restriction $x < x_{0}$ , $x$

is always smaller than $x_{0}$ . Then we write $x \rightarrow x_{0}-0$ or $x \rightarrow x_{0} -$ . Similarly, if $x$

appraoches $x_{0}$ taking larger value than $x_{0}$ , then we write $x \rightarrow x_{0}+0$ or $x \rightarrow x_{0} +$ .

Left-hand limit If $f(x)$ approaches $l$ as $x \rightarrow x_{0}-0$ , then we write

$\displaystyle \lim_{x \rightarrow x_{0}-0}f(x) = l$ or $\displaystyle f(x_{0}-0) = l$

and call

, left-hand limit.

Right-hand limit If $f(x)$ approaches $l$ as $x \rightarrow x_{0}+0$ , then we write

$\displaystyle \lim_{x \rightarrow x_{0}+0}f(x) = l$ or $\displaystyle f(x_{0}+0) = l$

and call

, right-hand limit

Example 1..25 Find the right-hand limit and the left-hand limit of the function at 0. $\displaystyle{f(x) = \left\{\begin{array}{rl} x^{2}+1,& x < 0\\ 0,&x = 0\\ x^{2} - 1,& x > 0 \end{array}\right.}$

SOLUTION When $x$

approaches 0 fram the left, $x$

is always smaller than 0. Thus $x < 0$

and $f(x) = x^2 + 1 \rightarrow 1$ . Therefore,

$\displaystyle \lim_{x \rightarrow 0-} f(x) = 1$

On the other hand, if $x$

approaches 0 from the right, then $x$

is always larger than 0. Thus $x > 0$

and $f(x) = x^2 - 1 \rightarrow -1$ . Therefore,

$\displaystyle \lim_{x \rightarrow 0+} f(x) = -1 \ensuremath{ \blacksquare}$

**Figure 1.40:** Example1-25
$% latex2html id marker 39931 \includegraphics[width=3.5cm]{SOFTFIG-1/reidai1-25_gr3.eps}$

Exercise 1..25 Find the right-hand limit and left-hand limit of the following function $\displaystyle{f(x) = \frac{\vert x\vert}{x}}$ at 0.

**Figure 1.41:** Exercise1-25
$% latex2html id marker 39942 \includegraphics[width=3.5cm]{SOFTFIG-1/enshu1-25_gr3.eps}$

SOLUTION Note that for $x < 0$ we have $\vert x\vert = -x$ .

$\displaystyle \lim_{x \rightarrow 0-}\frac{\vert x\vert}{x} = \lim_{x \rightarrow 0-}\frac{-x}{x} = -1$

$\displaystyle \lim_{x \rightarrow 0+}\frac{\vert x\vert}{x} = \lim_{x \rightarrow 0+}\frac{x}{x} = 1\ensuremath{ \blacksquare}$

Existence of a limit

Theorem 1..8 $\displaystyle{\lim_{x \rightarrow x_{0}}f(x) = l}$ iff $\displaystyle{\lim_{x \rightarrow x_{0}-0}f(x) = l}$ and $\displaystyle{\lim_{x \rightarrow x_{0}+0}f(x) = l}$ .

Non-existence of a limit To show non-existence of a limit, it is enough to show $\displaystyle{\lim_{x \rightarrow x_{0}-0}f(x) \neq \lim_{x \rightarrow x_{0}+0}f(x)}$ .

NOTE The existence of $f(x_{0}-0)$ and the existence of $f(x_{0}+0)$ , and equality of their values is suffice to say the existence of a limit of $f(x)$ . Otherwise, no limit exists.

Example 1..26 Find the limit of the following functions. $\displaystyle{1. \lim_{x \rightarrow 2}\frac{1}{x-2}}$ $\displaystyle{2. \lim_{x \rightarrow 0}\frac{\vert x\vert}{x}}$

SOLUTION 1.

$\displaystyle \lim_{x \to 2-}\frac{1}{x-2} = -\infty$

$\displaystyle \lim_{x \to 2+}\frac{1}{x-2} = \infty$

Thus, no limit exists $\blacksquare$

Check to see the existence of the right-hand limit and left-hand limit.

2. Since the right-hand limit is not equql to the left-hand limit, by Exercise, we have $\lim_{x \rightarrow 0}\frac{\vert x\vert}{x}$ $\blacksquare$

Exercise 1..26 Find the limit of $\displaystyle{\lim_{x \rightarrow 2}\frac{2-x}{\sqrt{(2-x)^2}}}$ .

Recall that $\sqrt{x^2} = \vert x\vert$ . Note that $x < 2$ implies $2 - x > 0$ which in turn implies $\sqrt{(2-x)^2} = 2-x$ . Note also that $x > 2$ implies $2- x < 0$ which in turn implies $\sqrt{(2-x)^2} = -(2-x)$ .

SOLUTION

$\displaystyle \lim_{x \rightarrow 2-}\frac{2-x}{\sqrt{(2-x)^2}} = \lim_{x \rightarrow 2-}\frac{2-x}{2-x} = 1$

$\displaystyle \lim_{x \rightarrow 2+}\frac{2-x}{\sqrt{(2-x)^2}} = \lim_{x \rightarrow 2+}\frac{2-x}{-(2-x)} = -1$

Thus no limit exists $\blacksquare$

Exercise A

1.

Find the limit of the following functions:

(a) $\displaystyle{\lim_{x \rightarrow 0}(2x -1)}$ (b) $\displaystyle{\lim_{x \rightarrow 1}\frac{x}{x + 1}}$ (c) $\displaystyle{\lim_{x \rightarrow 0}\frac{x^{2}(x + 1)}{2x}}$ (d) $\displaystyle{\lim_{x \rightarrow 0}\frac{x(x+1)}{2x}}$

(e) $\displaystyle{\lim_{x \rightarrow 9}\frac{x - 9}{\sqrt{x} - 3}}$ (f) $\displaystyle{\lim_{x \rightarrow 3}\frac{x^{2} - x- 6}{x-3}}$ (g) $\displaystyle{\lim_{x \rightarrow 4}\frac{(x^{2} - 3x - 4)^{2}}{x-4}}$

(h) $\displaystyle{\lim_{t \rightarrow 0}\frac{1}{t}\left(\frac{1}{t+3} - \frac{1}{3}\right)}$ (i) $\displaystyle{\lim_{x \rightarrow 0}x\left(1 - \frac{x + 1}{x}\right)}$

2.

Using $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{x}}{x} = 1}$ , find the limit of the following functions:

(a) $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{2x}}{\sin{3x}}}$ (b) $\displaystyle{\lim_{x \rightarrow 0}\frac{\cos^{2}{x} - 1}{x^{2}}}$ (c) $\displaystyle{\lim_{x \rightarrow 0}\frac{\cos{x} - 1}{x}}$ (d) $\displaystyle{\lim_{x \rightarrow \pi}\frac{\sin{x}}{x - \pi}}$

Exercise B

1.

Find the limit of the following functions:

(a) $\displaystyle{\lim_{x \rightarrow 2}(x^2 + 4x)}$ (b) $\displaystyle{\lim_{x \rightarrow 2}\frac{x^2 - 3x + 2}{x^2 -4}}$ (c) $\displaystyle{\lim_{x \rightarrow 0}\frac{2x^4 - 6x^3 + x^2 + 2}{x - 1}}$

(d) $\displaystyle{\lim_{x \rightarrow 1}\frac{2x^4 - 6x^3 + x^2 + 3}{x - 1}}$ (e) $\displaystyle{\lim_{x \rightarrow 0}\frac{\sqrt{1+x} - \sqrt{1 - x}}{x}}$

(f) $\displaystyle{\lim_{x \rightarrow 0}\frac{(1+x)^{1/3} - (1 - x)^{1/3}}{x}}$ (g) $\displaystyle{\lim_{x \rightarrow 2} f(x)}$ , ?????? $\displaystyle{f(x) = \left\{\begin{array}{cl} x^2, & x \neq 2\\ 3, & x = 2 \end{array} \right.}$

2.

Using $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{x}}{x} = 1}$ , find the limit of the followings:

(a) $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin{3x}}{x}}$ (b) $\displaystyle{\lim_{x \rightarrow 0}\frac{2x}{\sin{x}}}$ (c) $\displaystyle{\lim_{x \rightarrow 0}\frac{\sin^{-1}{x}}{x}}$ (d) $\displaystyle{\lim_{x \rightarrow 0}x \sin\frac{1}{x}}$

3.

Show that $\lim_{x \rightarrow a}\vert f(x)\vert = 0$ implies $\lim_{x \rightarrow a}f(x) = 0$ .