Limits of Polynomial, Rational and Irrational Functions

Lessons

Senior 4 · Mathematics

Limits of Polynomial, Rational and Irrational Functions

4 lessons

Lesson 2 of 4

One-sided limits, infinite limits and limits at infinity

25 min

Approaching from one side, and running off to infinity

In Unit 6 the image distance of a lens with focal length 10 cm was $v(u) = \frac{10u}{u - 10}$ . Move the object towards the focal point from outside ( $u \to 10$ , $u \gt 10$ ) and $v$ grows without bound; move the object far away ( $u \to +\infty$ ) and $v$ settles towards 10 cm. These are two new kinds of limit: an infinite limit and a limit at infinity. Both need a third idea first: approaching a point from one side only.

Definition

Left-hand limit: $\displaystyle\lim_{x \to a^-} f(x) = L$ means $f (x)$ is as close to $L$ as we like for all $x \lt a$ close enough to $a$ .

Right-hand limit: $\displaystyle\lim_{x \to a^+} f(x) = L$ , the same with $x \gt a$ .

Example: for $f(x) = \frac{|x - 2|}{x - 2}$ , $f (x) = - 1$ when $x \lt 2$ and $f (x) = 1$ when $x \gt 2$ , so $\displaystyle\lim_{x \to 2^-} f(x) = -1$ and $\displaystyle\lim_{x \to 2^+} f(x) = 1$ .

Important

Existence of a limit. $\displaystyle\lim_{x \to a} f(x) = L$ if and only if both one-sided limits exist and $\displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$ .

So $\frac{|x - 2|}{x - 2}$ has no limit at 2: its one-sided limits differ.

$y = \frac{|x - 2|}{x - 2}$ jumps at $x = 2$ : the left-hand limit is $- 1$ , the right-hand limit is $1$ , so the limit at 2 does not exist (and $f (2)$ is undefined).

Worked example 1: making a limit exist

$f(x) = \begin{cases} x^2 + 1 & x \lt 1 \\ 3x - k & x \ge 1 \end{cases}$ . Find $k$ so that $\displaystyle\lim_{x \to 1} f(x)$ exists.

Solution $\displaystyle\lim_{x \to 1^-} f(x) = 1 + 1 = 2$ and $\displaystyle\lim_{x \to 1^+} f(x) = 3 - k$ . The limit exists when $3 - k = 2$ , that is $k = 1$ ; then the limit is 2.

Definition

Infinite limit: $\displaystyle\lim_{x \to a} f(x) = +\infty$ means $f (x)$ becomes larger than any number $M$ , however large, for all $x \ne a$ close enough to $a$ . Similarly for $-\infty$ and for one-sided limits.

If $\displaystyle\lim_{x \to a^{\pm}} f(x) = \pm\infty$ (any combination), the line $x = a$ is a vertical asymptote of the graph.

Important

Quotient rule for infinite limits. If the numerator tends to a number $L \ne 0$ and the denominator tends to 0, the quotient tends to $+\infty$ or $-\infty$ . The sign comes from the signs of the numerator and of the denominator on that side; write $0^{+}$ or $0^{-}$ for a denominator that tends to 0 through positive or negative values.

Worked example 2: infinite limits

Find (a) $\displaystyle\lim_{x \to 2^-} \frac{1}{x - 2}$ and $\displaystyle\lim_{x \to 2^+} \frac{1}{x - 2}$ (b) $\displaystyle\lim_{x \to 1} \frac{x + 3}{(x - 1)^2}$ (c) $\displaystyle\lim_{u \to 10^+} \frac{10u}{u - 10}$ .

Solution

As $x \to 2^-$ , $x - 2 \to 0^-$ : $\frac{1}{0^-} \to -\infty$ . As $x \to 2^+$ , $x - 2 \to 0^+$ : $+\infty$ . The line $x = 2$ is a vertical asymptote.
Numerator $\to 4 \gt 0$ ; $(x - 1)^2 \to 0^+$ from both sides. So the limit is $+\infty$ (from both sides).
Numerator $\to 100$ ; $u - 10 \to 0^+$ . The limit is $+\infty$ : the image runs off to infinity as the object nears the focal point.

Definition

Limit at infinity: $\displaystyle\lim_{x \to +\infty} f(x) = L$ means $f (x)$ is as close to $L$ as we like for all $x$ large enough. Similarly for $x \to -\infty$ . Then the line $y = L$ is a horizontal asymptote.

Basic limits: for every whole number $n \ge 1$ , $\displaystyle\lim_{x \to \pm\infty} \frac{1}{x^n} = 0$ ; $\displaystyle\lim_{x \to +\infty} x^n = +\infty$ ; and $\displaystyle\lim_{x \to -\infty} x^n$ is $+\infty$ for even $n$ , $-\infty$ for odd $n$ .

$y = \frac{1}{x - 2}$ : $-\infty$ as $x \to 2^-$ , $+\infty$ as $x \to 2^+$ (vertical asymptote $x = 2$ ), and $0$ as $x \to \pm\infty$ (horizontal asymptote $y = 0$ ).

Important

Polynomials and rational functions at infinity. As $x \to \pm\infty$ , a polynomial behaves like its leading term. For a rational function, divide numerator and denominator by the highest power of $x$ in the denominator; the result depends on the degrees:

Degrees	$\displaystyle\lim_{x \to \pm\infty} \frac{a_nx^n + \cdots}{b_mx^m + \cdots}$	Example
$n \lt m$	$0$	$\frac{x + 1}{x^2 - 4} \to 0$
$n = m$	$\frac{a_n}{b_m}$ , the ratio of leading coefficients	$\frac{3x^2 - x + 2}{2x^2 + 5} \to \frac{3}{2}$
$n \gt m$	$+\infty$ or $-\infty$ , from the sign of $\frac{a_nx^n}{b_mx^m}$	$\frac{x^3 + 1}{2x - 1} \to +\infty$ as $x \to -\infty$

Worked example 3: rational and irrational functions at infinity

Find (a) $\displaystyle\lim_{x \to +\infty} \frac{3x^2 - x + 2}{2x^2 + 5}$ (b) $\displaystyle\lim_{x \to -\infty} (2x^3 - 5x^2 + 1)$ (c) $\displaystyle\lim_{x \to +\infty} \frac{\sqrt{x^2 + 1}}{2x - 3}$ and $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{x^2 + 1}}{2x - 3}$ .

Solution

Divide by $x^{2}$ : $\frac{3 - \frac{1}{x} + \frac{2}{x^2}}{2 + \frac{5}{x^2}} \to \frac{3 - 0 + 0}{2 + 0} = \frac{3}{2}$ .
$2x^3 - 5x^2 + 1 = x^3\left(2 - \frac{5}{x} + \frac{1}{x^3}\right)$ , with $x^3 \to -\infty$ and the bracket $\to 2$ : the limit is $-\infty$ .
$\sqrt{x^2 + 1} = |x|\sqrt{1 + \frac{1}{x^2}}$ . For $x \gt 0$ , $∣ x ∣ = x$ : $\frac{x\sqrt{1 + \frac{1}{x^2}}}{x\left(2 - \frac{3}{x}\right)} \to \frac{1}{2}$ . For $x \lt 0$ , $∣ x ∣ = - x$ : the same working gives $-\frac{1}{2}$ . The graph has two different horizontal asymptotes, $y = \frac{1}{2}$ and $y = -\frac{1}{2}$ .

Warning

$\sqrt{x^2} = |x|$ , not $x$ . When $x \to -\infty$ , $\sqrt{x^2} = -x$ . Forgetting this is the most common error with irrational functions at $-\infty$ , and it gives the wrong sign.

Tip

With infinite limits, keep a small "arithmetic of infinity" in mind: $L + \infty = \infty$ ; $L \times \infty = \pm\infty$ for $L \ne 0$ ; $\frac{L}{\infty} = 0$ ; $\frac{L}{0^{\pm}} = \pm\infty$ for $L \gt 0$ . The forms $\infty - \infty$ , $0 \times \infty$ , $\frac{\infty}{\infty}$ and $\frac{0}{0}$ are not covered: they are the indeterminate forms of Lesson 3.

The Organic Chemistry Tutor: Limits at infinity and horizontal asymptotes (about 19 min). Watch how the degrees of the numerator and the denominator decide the limit, and how a square root needs care when x tends to minus infinity.

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Practice

Show the sign analysis in every infinite limit.

For $f(x) = \frac{|x + 1|}{x + 1}$ , find the one-sided limits at $- 1$ . Does $\displaystyle\lim_{x \to -1} f(x)$ exist?
$g(x) = \begin{cases} 2x + m & x \lt 2 \\ x^2 - 1 & x \ge 2 \end{cases}$ . Find $m$ so that $\displaystyle\lim_{x \to 2} g(x)$ exists.
Find $\displaystyle\lim_{x \to 3^-} \frac{2x}{x - 3}$ and $\displaystyle\lim_{x \to 3^+} \frac{2x}{x - 3}$ .
Find $\displaystyle\lim_{x \to -2} \frac{x - 1}{(x + 2)^2}$ .
Find $\displaystyle\lim_{x \to +\infty} \frac{5x^3 - 2x}{x^3 + 4x^2 - 1}$ and $\displaystyle\lim_{x \to +\infty} \frac{2x + 7}{x^2 + 1}$ .
Find $\displaystyle\lim_{x \to -\infty} (x^4 - 3x^3 + 2)$ and $\displaystyle\lim_{x \to -\infty} \frac{x^2 + 1}{x - 2}$ .
Find $\displaystyle\lim_{x \to +\infty} \frac{\sqrt{4x^2 + x}}{x + 1}$ and $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{4x^2 + x}}{x + 1}$ .
For the lens $v(u) = \frac{10u}{u - 10}$ , find $\displaystyle\lim_{u \to +\infty} v(u)$ and explain what it means.

Show answersHide answers

Left limit $- 1$ , right limit $1$ ; different, so no limit.
Left: $4 + m$ ; right: $3$ . So $m = - 1$ .
Numerator $\to 6$ : $\frac{6}{0^-} = -\infty$ and $\frac{6}{0^+} = +\infty$ .
Numerator $\to -3$ , denominator $\to 0^+$ : $-\infty$ .
$5$ (equal degrees); $0$ (lower degree on top).
$+\infty$ (leading term $x^{4}$ ); $-\infty$ (behaves like $\frac{x^2}{x} = x$ ).
$\sqrt{4x^2 + x} = |x|\sqrt{4 + \frac{1}{x}}$ : $+\infty$ : $2$ ; $-\infty$ : $- 2$ .
Divide by $u$ : $\frac{10}{1 - \frac{10}{u}} \to 10$ . A very distant object forms its image at the focal point, 10 cm from the lens.

Quick check

For f(x) = 1 ÷ (x − 2), match each limit with its value.

Tap one on the left, then its partner on the right.

Key points

$\lim_{x \to a} f(x)$ exists exactly when both one-sided limits exist and are equal.
$\frac{L}{0^{\pm}}$ with $L \ne 0$ gives $\pm\infty$ : find the sign on each side; $x = a$ is then a vertical asymptote.
At $\pm\infty$ : a polynomial behaves like its leading term; for a rational function compare degrees (0, ratio of leading coefficients, or $\pm\infty$ ).
$\sqrt{x^2} = |x|$ : at $-\infty$ it equals $- x$ .

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Limits at Infinity & Horizontal Asymptotes, The Organic Chemistry Tutor, embedded from YouTube.

1Approaching from one side, and running off to infinity

Approaching from one side, and running off to infinity