Limits of Polynomial, Rational and Irrational Functions

Lessons

Senior 4 · Mathematics

Limits of Polynomial, Rational and Irrational Functions

4 lessons

Lesson 4 of 4

Continuity, discontinuity and asymptotes

25 min

Can you draw it without lifting the pen?

The air temperature outside your classroom changes gradually during a day: if it rises from 16 °C to 25 °C, it passes through every temperature in between. A parking charge that goes up by a fixed amount each started hour does not: it jumps. The first kind of function is continuous; the second has discontinuities. With limits you can say exactly what "no jump, no hole, no break" means, find where a graph breaks, and describe the lines it approaches far away: its asymptotes.

Definition

Continuity at a point. A function $f$ is continuous at $a$ if

$f (a)$ is defined;
$\displaystyle\lim_{x \to a} f(x)$ exists;
$\displaystyle\lim_{x \to a} f(x) = f(a)$ .

$f$ is continuous on an interval if it is continuous at every point of it (at an endpoint, using the one-sided limit).

Example: $f (x) = x^{2} + 1$ is continuous at 2, since $\displaystyle\lim_{x \to 2} (x^2 + 1) = 5 = f(2)$ .

Important

Continuous functions. Polynomials are continuous on $\mathbb{R}$ ; rational and irrational functions are continuous at every point of their domain (inside it, and one-sidedly at its endpoints). Sums, differences, products, quotients (where the denominator is not 0) and composites of continuous functions are continuous.

Definition

Discontinuity at $a$ : a point where $f$ is not continuous. Three types:

Removable: $\displaystyle\lim_{x \to a} f(x) = L$ exists, but $f (a)$ is undefined or $f(a) \ne L$ . Defining (or redefining) $f (a) = L$ removes it; the new function is the continuous extension of $f$ .
Jump: both one-sided limits exist but are different.
Infinite: at least one one-sided limit is $\pm\infty$ (a vertical asymptote).

Removable (hole at $x = 2$ ), jump (at $x = 5$ ) and infinite (vertical asymptote at $x = 8$ ). Only the first can be repaired by choosing one value.

Worked example 1: classify and repair

Study the continuity of (a) $f(x) = \frac{x^2 - 9}{x - 3}$ at 3 (b) $g(x) = \frac{|x|}{x}$ at 0 (c) $h(x) = \frac{1}{x^2}$ at 0.

Solution

$f (3)$ is undefined, but $\displaystyle\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6$ . Removable. The continuous extension is $\tilde f(x) = \frac{x^2 - 9}{x - 3}$ for $x \ne 3$ , $\tilde f(3) = 6$ (in fact $\tilde f(x) = x + 3$ ).
Left limit $- 1$ , right limit $1$ : jump discontinuity, which cannot be removed.
$\displaystyle\lim_{x \to 0} \frac{1}{x^2} = +\infty$ : infinite discontinuity; $x = 0$ is a vertical asymptote.

Worked example 2: choose a constant

$f(x) = \begin{cases} \frac{\sqrt{x + 1} - 1}{x} & x \ne 0 \\ k & x = 0 \end{cases}$ . Find $k$ so that $f$ is continuous at 0.

Solution By the conjugate (Lesson 3), $\frac{\sqrt{x + 1} - 1}{x} = \frac{1}{\sqrt{x + 1} + 1} \to \frac{1}{2}$ . Continuity needs $f (0) =$ the limit, so $k = \frac{1}{2}$ .

Important

Intermediate value theorem. If $f$ is continuous on $[a, b]$ and $N$ lies between $f (a)$ and $f (b)$ , then $f (c) = N$ for at least one $c$ in $[a, b]$ . In particular, if $f (a)$ and $f (b)$ have opposite signs, the equation $f (x) = 0$ has a root between $a$ and $b$ .

Example: $f (x) = x^{3} + x - 1$ is a polynomial, so continuous; $f(0) = -1 \lt 0$ and $f(1) = 1 \gt 0$ , so $x^{3} + x - 1 = 0$ has a root between 0 and 1.

Asymptotes

An asymptote is a line that the curve approaches more and more closely as it runs off to infinity.

Definition

Vertical asymptote $x = a$ : $\displaystyle\lim_{x \to a^-} f(x)$ or $\displaystyle\lim_{x \to a^+} f(x)$ is $\pm\infty$ .
Horizontal asymptote $y = L$ : $\displaystyle\lim_{x \to +\infty} f(x) = L$ or $\displaystyle\lim_{x \to -\infty} f(x) = L$ .
Oblique asymptote $y = m x + c$ ( $m \ne 0$ ): $\displaystyle\lim_{x \to \pm\infty} [f(x) - (mx + c)] = 0$ . Then $m = \lim_{x \to \pm\infty} \frac{f(x)}{x}, \qquad c = \lim_{x \to \pm\infty} [f(x) - mx] .$

For a rational function whose numerator has degree exactly one more than its denominator, divide: the quotient $m x + c$ is the oblique asymptote.

Worked example 3: all the asymptotes of a rational function

Find the asymptotes of $f(x) = \frac{x^2 + x - 1}{x - 1}$ and the position of the curve relative to the oblique one.

Solution Divide: $x^{2} + x - 1 = (x - 1) (x + 2) + 1$ , so

f(x) = x + 2 + \frac{1}{x - 1} .

Vertical: as $x \to 1^\pm$ , $\frac{1}{x - 1} \to \pm\infty$ : $x = 1$ .
Oblique: $f(x) - (x + 2) = \frac{1}{x - 1} \to 0$ as $x \to \pm\infty$ : $y = x + 2$ . (No horizontal asymptote.)
Position: $f(x) - (x + 2) = \frac{1}{x - 1}$ is positive for $x \gt 1$ (curve above the line) and negative for $x \lt 1$ (curve below).

$f(x) = x + 2 + \frac{1}{x - 1}$ : vertical asymptote $x = 1$ , oblique asymptote $y = x + 2$ . The curve is above the oblique asymptote for $x \gt 1$ and below it for $x \lt 1$ .

Worked example 4: an irrational function

Find the oblique asymptotes of $f(x) = \sqrt{x^2 + 2x}$ (domain $(-\infty, -2] \cup [0, +\infty)$ ).

Solution $\sqrt{x^2 + 2x} = |x|\sqrt{1 + \frac{2}{x}}$ .

As $x \to +\infty$ : $m = \lim \frac{f(x)}{x} = \lim \sqrt{1 + \frac{2}{x}} = 1$ , and $c = \lim \left(\sqrt{x^2 + 2x} - x\right) = \lim \frac{2x}{\sqrt{x^2 + 2x} + x} = \lim \frac{2}{\sqrt{1 + \frac{2}{x}} + 1} = 1$ . Asymptote $y = x + 1$ .
As $x \to -\infty$ : $∣ x ∣ = - x$ , so $m = - 1$ , and the same method gives $c = - 1$ . Asymptote $y = - x - 1$ .

Warning

Not every zero of a denominator gives a vertical asymptote. At $x = 3$ , $\frac{x^2 - 9}{x - 3}$ has a hole, not an asymptote, because the limit there is finite. Always compute the limit before drawing a vertical asymptote.

Tip

A checklist for asymptotes: (1) vertical: the excluded values of the domain where the limit is infinite; (2) horizontal: $\lim_{x \to \pm\infty} f(x)$ , if finite; (3) oblique: if $\frac{f(x)}{x}$ tends to a non-zero number $m$ , find $c$ . Treat $+\infty$ and $-\infty$ separately when roots or absolute values appear.

The Organic Chemistry Tutor: Continuity, and point, jump and infinite discontinuities (about 14 min). Watch how each type of discontinuity looks on a graph, and which ones are removable.

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Practice

Questions 6 to 8 are about asymptotes.

Is $f(x) = \frac{x + 2}{x^2 - 4}$ continuous at $- 2$ ? At 2? Classify each discontinuity.
Find $k$ so that $f(x) = \begin{cases} kx + 1 & x \le 2 \\ x^2 - k & x \gt 2 \end{cases}$ is continuous at 2.
Give the continuous extension of $g(x) = \frac{x^3 - 1}{x - 1}$ at 1.
Show that $x^{3} - 4 x + 1 = 0$ has a root between 1 and 2.
Find the domain on which $h(x) = \frac{\sqrt{x + 3}}{x - 1}$ is continuous.
Find the vertical and horizontal asymptotes of $f(x) = \frac{2x + 1}{x + 3}$ .
Find the asymptotes of $f(x) = \frac{2x^2 - 3x + 1}{x + 1}$ .
Find the oblique asymptote of $f(x) = \sqrt{x^2 - 4x}$ as $x \to +\infty$ .

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Neither point is in the domain. At $- 2$ : $\frac{1}{x - 2} \to -\frac{1}{4}$ , removable. At 2: infinite (vertical asymptote).
Left value $2 k + 1$ , right limit $4 - k$ : $2k + 1 = 4 - k \Rightarrow k = 1$ .
$\frac{x^3 - 1}{x - 1} = x^2 + x + 1 \to 3$ : define $g (1) = 3$ .
$f(1) = -2 \lt 0$ , $f(2) = 1 \gt 0$ , and $f$ is a continuous polynomial, so by the intermediate value theorem there is a root in $(1, 2)$ .
$x \ge -3$ and $x \ne 1$ : $[-3, 1) \cup (1, +\infty)$ .
$x = - 3$ and $y = 2$ .
$2 x^{2} - 3 x + 1 = (x + 1) (2 x - 5) + 6$ : vertical $x = - 1$ , oblique $y = 2 x - 5$ .
$m = 1$ ; $c = \lim (\sqrt{x^2 - 4x} - x) = \lim \frac{-4x}{\sqrt{x^2 - 4x} + x} = -2$ : $y = x - 2$ .

Quick check

Put the three conditions for f to be continuous at a in the order they are checked.

Tap the cards in the right order.

Key points

Continuous at $a$ : $f (a)$ defined, the limit exists, and they are equal.
Discontinuities: removable (finite limit; repair with $f (a) = L$ ), jump (one-sided limits differ), infinite (vertical asymptote).
Intermediate value theorem: a continuous function that changes sign on $[a, b]$ has a root there.
Asymptotes: vertical where the limit is infinite; horizontal $y = \lim_{x \to \pm\infty} f(x)$ ; oblique $y = m x + c$ with $m = \lim \frac{f(x)}{x}$ , $c = \lim (f(x) - mx)$ .

1Can you draw it without lifting the pen?

2Asymptotes

Can you draw it without lifting the pen?

Asymptotes