26.1 Limits of Functions
AHL 5.12
- how to find limits of continuous functions by substitution, and the properties of limits
- how to find a limit of the form \(\frac{0}{0}\) by factorising and cancelling
- limits of composite functions
The idea of a limit, and how to estimate a limit from a table or a graph, are in 24.1 (SL).
The idea of a limit
The idea of a limit was introduced in Section 24.1: \(\displaystyle\lim_{x \to a} f(x) = L\) means that \(f(x)\) gets closer and closer to \(L\) when \(x\) gets closer and closer to \(a\) (from both sides, with \(x \neq a\)). The value \(f(a)\) itself does not matter. A limit does not exist if the two sides approach different values, or if \(f(x)\) becomes infinitely large.
Limits of functions
Limits of continuous functions
The graph of \(y = 2x - 1\) has no holes, jumps or breaks: it is continuous. As \(x \to 3\), the value of \(2x - 1\) gets closer to \(2(3) - 1 = 5\). So
\[ \lim_{x \to 3} (2x - 1) = 5 = f(3). \]
For a continuous function, the limit is simply the value of the function.
Limit of a continuous function Not in the formula booklet
If \(f\) is continuous at \(x = a\), then
\[ \lim_{x \to a} f(x) = f(a). \]
Polynomials, \(\sin x\), \(\cos x\), \(\mathrm{e}^x\) are continuous everywhere. Rational functions, \(\ln x\) and \(\sqrt{x}\) are continuous wherever they are defined.
- \(a\): the value that \(x\) approaches
- \(f(a)\): the value of \(f\) at \(x = a\)
Example 1 Find
(a) \(\displaystyle\lim_{x \to 4} (3x - 5)\)
(b) \(\displaystyle\lim_{x \to -2} (x^2 - x + 3)\).
Both are polynomials, so substitute.
(a) \(\displaystyle\lim_{x \to 4} (3x - 5) = 3(4) - 5 = 7\).
(b) \(\displaystyle\lim_{x \to -2} (x^2 - x + 3) = 4 + 2 + 3 = 9\).
Properties of limits
Properties of limits Not in the formula booklet
Let \(\displaystyle\lim_{x \to a} f(x) = L\) and \(\displaystyle\lim_{x \to a} g(x) = M\), and let \(k\) be a constant. Then
- \(\displaystyle\lim_{x \to a} k = k\)
- \(\displaystyle\lim_{x \to a} k\,f(x) = kL\)
- \(\displaystyle\lim_{x \to a} \big[f(x) \pm g(x)\big] = L \pm M\)
- \(\displaystyle\lim_{x \to a} \big[f(x)\,g(x)\big] = LM\)
- \(\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}\), provided \(M \neq 0\).
Example 2 Find
(a) \(\displaystyle\lim_{x \to 5} 7\)
(b) \(\displaystyle\lim_{x \to 2} \frac{x + 4}{x^2 + 5}\).
(a) The function is the constant \(7\), so the limit is \(7\).
(b) The limit of the denominator is \(2^2 + 5 = 9 \neq 0\), so we can use property 5:
\[ \lim_{x \to 2} \frac{x + 4}{x^2 + 5} = \frac{2 + 4}{4 + 5} = \frac{6}{9} = \frac{2}{3}. \]
Limits of the form \(\frac{0}{0}\)
A table works, but there is a quicker method. We cannot substitute \(x = 3\) into \(\dfrac{x^2 - 9}{x - 3}\). But the limit only uses values with \(x \neq 3\), so \(x - 3 \neq 0\) and we may cancel it:
\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x + 3)(x - 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 6. \]
Limits of the form \(\frac{0}{0}\)
If substitution gives \(\frac{0}{0}\), simplify first, then substitute:
- factorise and cancel the common factor (often \(x - a\)), or
- for square roots, multiply by the conjugate, e.g. \(\sqrt{u} - 3\) by \(\sqrt{u} + 3\).
Example 3 Find
(a) \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 5x + 6}\)
(b) \(\displaystyle\lim_{x \to 3} \frac{x^3 - 27}{x - 3}\)
(c) \(\displaystyle\lim_{h \to 0} \frac{(x + h)^2 - (x - h)^2}{h}\)
(d) \(\displaystyle\lim_{x \to 4} \frac{\sqrt{x + 5} - 3}{x - 4}\)
Each one gives \(\frac{0}{0}\) when we substitute, so we simplify first.
(a) \(\displaystyle\lim_{x \to 2} \frac{(x + 2)(x - 2)}{(x - 2)(x - 3)} = \lim_{x \to 2} \frac{x + 2}{x - 3} = \frac{4}{-1} = -4\).
(b) Use \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):
\[ \lim_{x \to 3} \frac{(x - 3)(x^2 + 3x + 9)}{x - 3} = \lim_{x \to 3} (x^2 + 3x + 9) = 9 + 9 + 9 = 27. \]
(c) Here \(h\) is the variable and \(x\) is a constant. Expand the top:
\[ (x + h)^2 - (x - h)^2 = (x^2 + 2xh + h^2) - (x^2 - 2xh + h^2) = 4xh. \]
So \(\displaystyle\lim_{h \to 0} \frac{4xh}{h} = \lim_{h \to 0} 4x = 4x\).
(d) Multiply the top and bottom by the conjugate \(\sqrt{x + 5} + 3\):
\[ \frac{\sqrt{x + 5} - 3}{x - 4} \times \frac{\sqrt{x + 5} + 3}{\sqrt{x + 5} + 3} = \frac{(x + 5) - 9}{(x - 4)\big(\sqrt{x + 5} + 3\big)} = \frac{1}{\sqrt{x + 5} + 3}. \]
So the limit is \(\dfrac{1}{\sqrt{9} + 3} = \dfrac{1}{6}\).
Limits of composite functions
If \(y = u^2\) and \(u = 3x + 1\), then \(y = (3x + 1)^2\) is a composite function of \(x\). In general, if \(y = f(u)\) and \(u = g(x)\), then \(y = f\big(g(x)\big)\).
Limit of a composite function Not in the formula booklet
If \(\displaystyle\lim_{x \to a} g(x) = M\) and \(f\) is continuous at \(u = M\), then
\[ \lim_{x \to a} f\big(g(x)\big) = f(M). \]
In words: find the limit of the inside first, then put it into the outside function.
- \(g\): the inside function
- \(f\): the outside function
- \(a\): the value that \(x\) approaches
- \(M\): the limit of the inside, \(\displaystyle\lim_{x \to a} g(x)\)
- \(u\): the input of the outside function \(f\)
Example 4 Let \(f(x) = 2x^2 - x + 1\) and \(g(x) = x^2 + 3x - 2\). Find \(\displaystyle\lim_{x \to 1} f\big(g(x)\big)\).
Inside: \(\displaystyle\lim_{x \to 1} g(x) = 1 + 3 - 2 = 2\).
\(f\) is a polynomial, so it is continuous at \(2\):
\[ \lim_{x \to 1} f\big(g(x)\big) = f(2) = 2(4) - 2 + 1 = 7. \]
Example 5 Find
(a) \(\displaystyle\lim_{x \to 3} \log(x^2 + x - 2)\)
(b) \(\displaystyle\lim_{x \to \frac{\pi}{4}} \cos\left(3x + \frac{\pi}{4}\right)\).
(a) Inside: \(x^2 + x - 2 \to 9 + 3 - 2 = 10\). \(\log\) is continuous at \(10\), so the limit is \(\log 10 = 1\).
(b) Inside: \(3x + \dfrac{\pi}{4} \to \dfrac{3\pi}{4} + \dfrac{\pi}{4} = \pi\). So the limit is \(\cos \pi = -1\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| “\(\frac{0}{0}\), so the limit does not exist.” | Simplify first, then substitute. | \(\frac{0}{0}\) only tells you that more work is needed. |
| \(\displaystyle\lim_{x \to a} f(x) = f(a)\) for every graph | Only when \(f\) is continuous at \(a\) | At a hole or a jump, \(f(a)\) can be different from the limit. |
| \(\displaystyle\frac{x^2 - 4}{x - 2} = x - 2\) | \(\dfrac{(x + 2)(x - 2)}{x - 2} = x + 2\) | Factorise fully before cancelling. |
| Using \(\dfrac{L}{M}\) when \(M = 0\) | Property 5 needs \(M \neq 0\) | If \(M = 0\), simplify first or look at both sides. |
Exercises
1 Find \(\displaystyle\lim_{x \to -1} (4x^3 - 2x + 5)\).
Polynomial, so substitute: \(4(-1) + 2 + 5 = 3\).
2 Find \(\displaystyle\lim_{x \to 2} \frac{x^2 + 5}{3x - 1}\).
The denominator tends to \(5 \neq 0\), so substitute: \(\dfrac{4 + 5}{6 - 1} = \dfrac{9}{5}\).
3 Find \(\displaystyle\lim_{x \to -3} \frac{x^2 + x - 6}{x + 3}\).
\(\dfrac{(x + 3)(x - 2)}{x + 3} = x - 2 \to -3 - 2 = -5\).
4 Find \(\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}\).
\(\dfrac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \dfrac{x^2 + x + 1}{x + 1} \to \dfrac{3}{2}\).
5 Find \(\displaystyle\lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}\).
\((3 + h)^2 - 9 = 6h + h^2\), so \(\dfrac{6h + h^2}{h} = 6 + h \to 6\).
6 Find \(\displaystyle\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}\).
\(x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)\), so the fraction is \(\dfrac{1}{\sqrt{x} + 3} \to \dfrac{1}{6}\).
7 Let \(f(u) = u^2 - 3u\) and \(g(x) = \dfrac{x^2 - 4}{x - 2}\). Find \(\displaystyle\lim_{x \to 2} f\big(g(x)\big)\).
\(\displaystyle\lim_{x \to 2} g(x) = \lim_{x \to 2} (x + 2) = 4\). \(f\) is continuous, so the limit is \(f(4) = 16 - 12 = 4\).
8 Find
(a) \(\displaystyle\lim_{x \to 2} \ln(3x - 5)\)
(b) \(\displaystyle\lim_{x \to \frac{\pi}{2}} \sin\left(\frac{x}{3} + \frac{\pi}{3}\right)\).
(a) \(3x - 5 \to 1\), so the limit is \(\ln 1 = 0\).
(b) \(\dfrac{x}{3} + \dfrac{\pi}{3} \to \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}\), so the limit is \(\sin\dfrac{\pi}{2} = 1\).