CALCULUS · HL
Limits
What a function heads towards.
Section 1 of 7
The idea of a limit
A limit asks: what value does an expression head towards as $x$ approaches some number?
(i) As $x \to \infty$
| $x$ | $\dfrac{1}{x}$ |
| 1 | 1 |
| 10 | 0.1 |
| 100 | 0.01 |
| 1000 | 0.001 |
| ↓ | ↓ |
| ∞ | 0 |
$\lim_{x \to \infty} \dfrac{1}{x} = 0$
(ii) As $x \to 0$
| $x$ | $\dfrac{1}{x}$ |
| 1 | 1 |
| 0.1 | 10 |
| 0.01 | 100 |
| 0.001 | 1000 |
| ↓ | ↓ |
| 0 | ∞ |
$\lim_{x \to 0} \dfrac{1}{x} = \infty$
YOU TRY · 1
State $\lim_{x \to \infty} \dfrac{1}{x^2}$.
As $x$ grows, the bottom grows even faster.
$0$
YOU TRY · 2
As $x \to 0$, what does $\dfrac{1}{x}$ do?
Dividing by a tiny number gives a huge number.
It heads to $\infty$ (grows without bound).
Section 2 of 7
Limits by substitution
If subbing the value straight in gives a number, that number is the limit.
(i) $5x - x^2$
$\lim_{x\to 1}(5x - x^2) = 5(1) - 1 = 4$
$\lim_{x\to 2}(5x - x^2) = 10 - 4 = 6$
$\lim_{x\to 3}(5x - x^2) = 15 - 9 = 6$
$\lim_{x\to 4}(5x - x^2) = 20 - 16 = 4$
(ii) $\lim_{x\to 5}(3x+1)$
Sub in $x = 5$:
$3(5) + 1 = 16$
YOU TRY · 3
Find $\lim_{x\to 2}(x^2 + 3x - 1)$.
Just sub $x = 2$.
$4 + 6 - 1$
$= 9$
$9$
YOU TRY · 4
Find $\lim_{x\to -1}(2x^2 - x + 4)$.
Sub $x = -1$. Mind the signs.
$2(1) - (-1) + 4 = 2 + 1 + 4$
$= 7$
$7$
Section 3 of 7
The $\dfrac{0}{0}$ problem
Sometimes subbing in gives $\dfrac{0}{0}$ — that's indeterminate. It needs work before you can read it off.
Contrast: $\dfrac{0}{5} = 0$ but $\dfrac{5}{0} = \infty$. $\dfrac{0}{0}$ is neither.
(i) $\lim_{x\to 3}\dfrac{x^2-9}{x-3}$
Sub $x=3$: $\dfrac{0}{0}$ indeterminate
Factor the top — difference of two squares:
$\lim_{x\to 3}\dfrac{(x-3)(x+3)}{x-3}$
Cancel $(x-3)$: $\lim_{x\to 3}(x+3)$
$= 3 + 3 = 6$
(ii) $\lim_{x\to 5}\dfrac{x-5}{x^2-5x}$
Sub $x=5$: $\dfrac{0}{0}$
Factor the bottom: $\lim_{x\to 5}\dfrac{x-5}{x(x-5)}$
Cancel $(x-5)$: $\lim_{x\to 5}\dfrac{1}{x}$
$= \dfrac{1}{5}$
(iii) expanded top — factor first
$\lim_{x\to 3}\dfrac{x^2-4x+3}{x-3} = \lim_{x\to 3}\dfrac{(x-1)(x-3)}{x-3}$
Cancel $(x-3)$: $\lim_{x\to 3}(x-1)$
$= 3 - 1 = 2$
YOU TRY · 5
Find $\lim_{x\to 2}\dfrac{x^2-4}{x-2}$.
Top is a difference of two squares.
$\dfrac{(x-2)(x+2)}{x-2} = x+2$
$= 4$
$4$
YOU TRY · 6
Find $\lim_{x\to 0}\dfrac{x^2+3x}{x}$.
Factor an $x$ out of the top.
$\dfrac{x(x+3)}{x} = x+3$
$= 3$
$3$
Section 4 of 7
Surds — multiply by the conjugate
A surd that leaves you with $\dfrac{0}{0}$? Multiply top and bottom by the conjugate to clear it.
(i) $\lim_{x\to 4}\dfrac{x-4}{\sqrt{x}-2}$
Sub $x=4$: $\dfrac{0}{\sqrt{4}-2} = \dfrac{0}{0}$
Multiply by $\dfrac{\sqrt{x}+2}{\sqrt{x}+2}$:
$\lim_{x\to 4}\dfrac{(x-4)(\sqrt{x}+2)}{x-4}$
Cancel $(x-4)$: $\lim_{x\to 4}(\sqrt{x}+2)$
$= \sqrt{4} + 2 = 4$
(ii) $\lim_{x\to 9}\dfrac{\sqrt{x}-3}{x-9}$
Sub $x=9$: $\dfrac{0}{0}$
Multiply by $\dfrac{\sqrt{x}+3}{\sqrt{x}+3}$:
$\lim_{x\to 9}\dfrac{x-9}{(x-9)(\sqrt{x}+3)}$
Cancel $(x-9)$: $\lim_{x\to 9}\dfrac{1}{\sqrt{x}+3}$
$= \dfrac{1}{3+3} = \dfrac{1}{6}$
Same result by difference of two squares: $x - 9 = (\sqrt{x})^2 - 3^2 = (\sqrt{x}-3)(\sqrt{x}+3)$.
YOU TRY · 7
Find $\lim_{x\to 16}\dfrac{x-16}{\sqrt{x}-4}$.
Multiply by $\dfrac{\sqrt{x}+4}{\sqrt{x}+4}$.
Cancel $(x-16)$: $\;\sqrt{x}+4$
$= \sqrt{16}+4 = 8$
$8$
YOU TRY · 8
Find $\lim_{x\to 1}\dfrac{\sqrt{x}-1}{x-1}$.
Multiply by $\dfrac{\sqrt{x}+1}{\sqrt{x}+1}$.
$\dfrac{x-1}{(x-1)(\sqrt{x}+1)} = \dfrac{1}{\sqrt{x}+1}$
$= \dfrac{1}{2}$
$\dfrac{1}{2}$
Section 5 of 7
Limits at infinity
Rule
→As $x \to \infty$, divide top and bottom by the highest power of $x$. Then every $\dfrac{a}{x^n} \to 0$.
(i) $\lim_{x\to\infty}\dfrac{2x+1}{x-3}$
Subbing $\infty$ gives $\dfrac{\infty}{\infty}$ — divide by $x$:
$\lim_{x\to\infty}\dfrac{2 + \dfrac{1}{x}}{1 - \dfrac{3}{x}}$
$= \dfrac{2+0}{1-0} = 2$
(ii) $\lim_{x\to\infty}\dfrac{3x+2}{5x-7}$
$\lim_{x\to\infty}\dfrac{3 + \dfrac{2}{x}}{5 - \dfrac{7}{x}}$
$= \dfrac{3}{5}$
(iii) $\lim_{x\to\infty}\dfrac{5x^2-6x}{7-3x^2}$
Highest power is $x^2$ — divide by $x^2$:
$\lim_{x\to\infty}\dfrac{5 - \dfrac{6}{x}}{\dfrac{7}{x^2} - 3}$
$= \dfrac{5 - 0}{0 - 3} = -\dfrac{5}{3}$
YOU TRY · 9
Find $\lim_{x\to\infty}\dfrac{4x-1}{2x+5}$.
Divide top and bottom by $x$.
$\dfrac{4 - \tfrac{1}{x}}{2 + \tfrac{5}{x}} \to \dfrac{4}{2}$
$= 2$
$2$
YOU TRY · 10
Find $\lim_{x\to\infty}\dfrac{x^2+1}{3x^2-x}$.
Highest power is $x^2$ — divide by it.
$\dfrac{1 + \tfrac{1}{x^2}}{3 - \tfrac{1}{x}} \to \dfrac{1}{3}$
$= \dfrac{1}{3}$
$\dfrac{1}{3}$
Section 6 of 7
Powers and $r^n$
$\lim_{x\to\infty}\dfrac{1}{x} = 0$
$\lim_{x\to\infty}\dfrac{1}{x^n} = 0$
$\lim_{n\to\infty} 2^n = \infty$ (grows)
$\lim_{n\to\infty} \left(\dfrac{1}{2}\right)^n = 0$ (shrinks)
Key fact
•$\lim_{n\to\infty} r^n = 0$ if $|r| < 1$.
•$|r| < 1 \Rightarrow -1 < r < 1$ — so $r$ is a proper fraction $\dfrac{a}{b}$ with $a < b$.
YOU TRY · 11
Find $\lim_{n\to\infty} \left(\dfrac{1}{3}\right)^n$.
Is $|r| < 1$?
$\left|\dfrac{1}{3}\right| < 1$, so $r^n \to 0$.
$0$
YOU TRY · 12
Does $\lim_{n\to\infty} \left(\dfrac{3}{2}\right)^n$ settle on a finite value?
Check whether $|r| < 1$.
$\left|\dfrac{3}{2}\right| > 1$, so it grows.
No — it heads to $\infty$.
Section 7 of 7
Recap — the toolkit
Must learn
1.$\lim_{x\to\infty}\dfrac{1}{x} = 0$ and $\lim_{x\to 0}\dfrac{1}{x} = \infty$.
2.First try: substitute. A number is the answer.
3.$\dfrac{0}{0}$ → factor and cancel, then substitute.
4.A surd giving $\dfrac{0}{0}$ → multiply by the conjugate.
5.$x \to \infty$ → divide by the highest power of $x$.
6.$\lim_{n\to\infty} r^n = 0$ when $|r| < 1$.
End of lesson
Limits — HL · Mathslive.ie