One sided limit

When $x$ 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

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

and call $l$, left-hand limit.

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

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

and call $l$, right-hand limit

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

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,

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

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,

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

Figure 1.40: Example1-25

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

Figure 1.41: Exercise1-25

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

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

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

Existence of a limit

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

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

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. $${1. \lim_{x \rightarrow 2}\frac{1}{x-2}}$$ $${2. \lim_{x \rightarrow 0}\frac{\vert x\vert}{x}}$$

SOLUTION 1.

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

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

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 $${\lim_{x \rightarrow 2}\frac{2-x}{\sqrt{(2-x)^2}}}$$.

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

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

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

Thus no limit exists $\blacksquare$

Exercise A

1.

Find the limit of the following functions:

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

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

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

2.

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

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

Exercise B

1.

Find the limit of the following functions:

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

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

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

2.

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

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

3.

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