Exam practice

Infinity (∞) means without end or without bound; it is an idea, not a number. Yes, the integers …, −2, −1, 0, 1, 2, … extend to infinity in both directions: there is no largest integer (add 1 to any integer and you get a bigger one) and no smallest (subtract 1). They run towards + ∞ and −∞.

∞ + ∞ = ∞; (−∞) + (−∞) = −∞.

(+ ∞) × (+ ∞) = ∞; (−∞) × (−∞) = ∞; (−∞) × (∞) = −∞.

• i.−∞
• ii.−∞
• iii.∞
• iv.∞ (taking away −∞ is adding ∞)
• v.∞
• vi.−∞
• vii.−∞
• viii.∞ (negative times negative)

A fixed number can’t stop something that grows without end; only its sign, when multiplying, decides the direction. (For x = 0, 0 × ∞ has no single answer: that is why v to viii say x > 0 or x < 0.)

$f(0) = \frac{5}{0 - 1} =$ −5
$f(-1) = \frac{5}{1 - 1} = \frac{5}{0}$: undefined
$f(1) = \frac{5}{1 - 1} = \frac{5}{0}$: undefined
$f(2) = \frac{5}{4 - 1} =$ $\frac{5}{3}$

Where the bottom is 0: $x^2 - 1 = 0$, (x − 1)(x + 1) = 0, so x = 1 or x = −1.

$f(0) = \frac{1}{0 - 0 + 6} =$ $\frac{1}{6}$
$f(1) = \frac{1}{1 - 5 + 6} =$ $\frac{1}{2}$
$f(2) = \frac{1}{4 - 10 + 6} = \frac{1}{0}$: undefined
$f(3) = \frac{1}{9 - 15 + 6} = \frac{1}{0}$: undefined

$x^2 - 5x + 6 = 0$, (x − 2)(x − 3) = 0, so x = 2 or x = 3.

Each graph has two branches that shoot off up and down on either side of one vertical line, which they never touch; there the function has no value.

• a.x = 0 (the branches run along the Y-axis). This is the graph of $y = \frac{1}{x}$.
• b.x = 2 (the dashed line), as for $y = \frac{1}{x - 2}$.
• c.x = −1 (the dashed line), as for $y = \frac{1}{x + 1}$.

As the sides increase, the polygon gets as close to the circle as we like, so the shaded gap gets as small as we like. Its limiting value is 0.

Infinitely many: every triangle has midpoints, so a smaller one always fits inside.

Joining midpoints halves the side (the midpoint theorem): if the first side is a, the sides are $a, \frac{a}{2}, \frac{a}{4}, \frac{a}{8}, \ldots$, which head for 0.

The perimeters $3a, \frac{3a}{2}, \frac{3a}{4}, \ldots$ halve each time and head for 0.

Each triangle is a quarter of the one before (half the side, so $\frac{1}{4}$ the area): $A, \frac{A}{4}, \frac{A}{16}, \ldots$ head for 0.

Any set of circles with one center and radii shrinking by the same fraction will do: for example radii 4 cm, 2 cm, 1 cm, 0.5 cm (each half the one before), or 5 cm, 4 cm, 3.2 cm, 2.56 cm (each $\frac{4}{5}$). Draw them with a compass, keeping its point on the same center.

Infinitely many: a smaller circle always fits inside the last one.

With the first radius r, the radii are $r, \frac{3}{4}r, \left(\frac{3}{4}\right)^2 r, \ldots$, heading for 0. The circumferences $2\pi r, 2\pi \cdot \frac{3}{4}r, \ldots$ head for 0: the circles shrink to the center point.

The areas $\pi r^2, \pi \left(\frac{3}{4}r\right)^2, \ldots$ head for 0 as well.

As x increases, y decreases, falling quickly at first and then more and more slowly; the curve gets closer and closer to the X-axis. But it never reaches 0: there is no x for which y = 0. We say y → 0 as x → ∞. (A curve like this is $y = \frac{2}{x}$: $\frac{2}{x}$ is never 0, however large x is.)

0.1, 0.01, 0.001, … head for 0: $\lim_{n \to \infty} \frac{1}{10^n} = 0$.

n = 1, 10, 100, 1000: $\frac{1}{3} \approx 0.333$, $\frac{10}{102} \approx 0.098$, $\frac{100}{10002} \approx 0.0100$, $\frac{1000}{1000002} \approx 0.0010$. The bottom grows much faster than the top (it is more than n times the top), so the terms head for 0.

$\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots, \frac{1}{102}, \ldots$: the bottom grows without end, so the terms head for 0.

$\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \ldots, \frac{1}{2000}, \ldots$ head for 0.

1, 4, 9, 16, …, 10,000, … grow without end: f(n) → ∞. There is no real number the terms settle on; we write $\lim_{n \to \infty} n^2 = \infty$, meaning the sequence grows without bound.

2, 4, 6, 8, … grow without end: f(n) → ∞, no real-number limit (written $\lim_{n \to \infty} 2n = \infty$).

0

The part after 7 shrinks to 0: the limit is 7.

The gap below 3 is 0.1, 0.01, 0.001, …, heading for 0: the limit is 3.

Examples you may find: mandalas in paubha and thangka paintings (squares inside circles); rangoli and floor designs at festivals; wheels and hubcaps with spokes; the Ashoka chakra on India’s flag (24 spokes); a stop sign (an octagon); tiled patterns in temple windows and on the bottoms of baskets; coins with many-sided edges.

• a.Triangles, squares, pentagons, hexagons, octagons, and polygons with many sides; often also stars made from them.
• b.The circle: the more sides the polygons have, the closer they get to the circle they are drawn in.

The lengths are 10, 5, 2.5, 1.25, 0.625, 0.3125, … cm, each half the one before: $f(n) = \frac{10}{2^{n-1}}$. They head for 0: the limiting value is 0 cm. (Like Pasang’s race: the pieces get as small as we like.)

Infinitely many inner squares can be drawn.
The outer side is $\sqrt{16} = 4$ units. Each new side is $\frac{1}{\sqrt{2}}$ of the one before (as in Example 3), so the sides are $4, 2\sqrt{2}, 2, \sqrt{2}, 1, \frac{1}{\sqrt{2}}, \ldots$ units.

Each square has half the area of the one before: 16, 8, 4, 2, 1, $\frac{1}{2}$, … square units. The limiting value is 0.

4 times each side: $16, 8\sqrt{2}, 8, 4\sqrt{2}, 4, \ldots$ units (16, 11.31, 8, 5.66, 4, …). The limiting value is 0.

The sum is the series $16 + 8 + 4 + 2 + 1 + \frac{1}{2} + \cdots$. Its partial sums are 16, 24, 28, 30, 31, 31.5, 31.75, …, and each falls short of 32 by exactly the last area added (16, 8, 4, 2, …), a gap that heads for 0. So the sum, the limit of the partial sums, is 32 square units: twice the outer square.

x takes values as close to −3 as we like, from both sides, but never equal to −3: from the left −3.1, −3.01, −3.001, … ($x \to -3^-$) and from the right −2.9, −2.99, −2.999, … ($x \to -3^+$).

x approaches 5 from the left: it takes values a little less than 5, getting as close as we like, 4.9, 4.99, 4.999, …, but never 5.

As x gets as close to 2 as we like from both sides (never equal to 2), the value of f(x) gets as close to −5 as we like: the left-hand and right-hand limits at 2 are both −5.

As x approaches −2 from the left (−2.1, −2.01, −2.001, …), f(x) gets as close to 1 as we like: the left-hand limit at −2 is 1.

As x approaches −3 from the right (−2.9, −2.99, −2.999, …), f(x) gets as close to −4 as we like: the right-hand limit at −3 is −4.

$f(0) = \frac{0 - 9}{0 - 3} = \frac{-9}{-3} =$ 3
$f(1) = \frac{1 - 9}{1 - 3} = \frac{-8}{-2} =$ 4
$f(4) = \frac{4 - 9}{2 - 3} = \frac{-5}{-1} =$ 5
$f(9) = \frac{9 - 9}{3 - 3} =$ $\frac{0}{0}$, indeterminate.

x = 9, where top and bottom are both 0.

$f(0) = \frac{-16}{-4} =$ 4
$f(1) = \frac{1 - 16}{1 - 4} = \frac{-15}{-3} =$ 5
$f(3) = \frac{9 - 16}{3 - 4} = \frac{-7}{-1} =$ 7
$f(4) = \frac{16 - 16}{4 - 4} =$ $\frac{0}{0}$, indeterminate.

x = 4.

Both sides head for 5: $\lim_{x \to 3} (x + 2) =$ 5.

Both sides head for −7: $\lim_{x \to -2} (2x - 3) =$ −7.

Both sides head for 18: $\lim_{x \to 4} (x^2 + 2) =$ 18.

For x ≠ −3, $\frac{x^2 - 9}{x + 3} = \frac{(x + 3)(x - 3)}{x + 3} = x - 3$.

Both sides head for −6: $\lim_{x \to -3} \frac{x^2 - 9}{x + 3} =$ −6 (though f(−3) is $\frac{0}{0}$).

• i.$f(2) = \frac{4 - 4}{2 - 2} = \frac{0}{0}$, indeterminate.
• ii.f(x) = x + 2 for x ≠ 2: f(1.9) = 3.9, f(1.99) = 3.99, f(1.999) = 3.999, f(1.9999) = 3.9999. So $\lim_{x \to 2^-} f(x) =$ 4.
• iii.f(2.01) = 4.01, f(2.001) = 4.001, f(2.0001) = 4.0001, f(2.00001) = 4.00001. So $\lim_{x \to 2^+} f(x) =$ 4.
• iv.Both one-sided limits are 4, so $\lim_{x \to 2} f(x) =$ 4.

• i.$f(3) = \frac{9 - 9}{3 - 3} = \frac{0}{0}$, indeterminate.
• ii.f(x) = x + 3 for x ≠ 3: 5.9, 5.99, 5.999, 5.9999. So $\lim_{x \to 3^-} f(x) =$ 6.
• iii.6.01, 6.001, 6.0001, 6.00001. So $\lim_{x \to 3^+} f(x) =$ 6.
• iv.$\lim_{x \to 3} f(x) =$ 6.

Both sides head for 7: the limit is 7.

The limit is −9.

The limit is −8.

The limit is −1.

With $f(x) = 3x^2 + 1$:

The limit is 13.

The limit is 5.

Both sides head for 0.4: the limit is $\frac{2}{5}$ (and $\frac{3 - 1}{3 + 2} = \frac{2}{5}$).

From the left the values grow without end (→ ∞), from the right they fall without end (→ −∞). They head for no number, so the limit does not exist.

Both sides head for 4: the limit is 4 (and $\frac{-3 - 1}{-3 + 2} = \frac{-4}{-1} = 4$).

From the left → −∞, from the right → ∞: the limit does not exist.

For x ≠ 3, $\frac{2x - 6}{x - 3} = \frac{2(x - 3)}{x - 3} = 2$.

The limit is 2 (though f(3) is $\frac{0}{0}$).

From the left → −∞, from the right → ∞: the limit does not exist.

For x ≠ 3 it equals x + 3.

The limit is 6.

The limit is 6. Check: $x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)$, so for x ≠ 9, $f(x) = \sqrt{x} + 3 \to 3 + 3 = 6$.

For x ≠ 4 it equals x + 4.

The limit is 8.

The limit is 0.8 = $\frac{4}{5}$. Check: $\frac{(x - 2)(x + 2)}{(x - 2)(x + 3)} = \frac{x + 2}{x + 3} \to \frac{4}{5}$.

Near −3 the top is close to 9 − 18 + 6 = −3 while the bottom is close to 0, so the values grow without end: towards −∞ from the left and ∞ from the right. The limit does not exist.
Some books give 0 as the answer here; 0 is the limit of $\frac{x^2 + 6x + 9}{x^2 - 9} = \frac{(x + 3)^2}{(x + 3)(x - 3)} = \frac{x + 3}{x - 3}$, with 9 in place of 6. Check which function your question gives.

The top is close to 2 and the bottom close to 0: → −∞ from the left, → ∞ from the right. The limit does not exist.

From both sides the values head for 3. Or by factorizing: $x^3 - 1 = (x - 1)(x^2 + x + 1)$, so for x ≠ 1, $f(x) = x^2 + x + 1 \to 1 + 1 + 1 = 3$. So $\lim_{x \to 1} \frac{x^3 - 1}{x - 1} = 3$. Proved.

From both sides the values head for 0.7. Here nothing is 0 at x = 3: the bottom is 10 ≠ 0, so the limit is just the value, $\frac{9 - 6 + 4}{9 + 1} = \frac{7}{10}$. So $\lim_{x \to 3} \frac{x^2 - 2x + 4}{x^2 + 1} = \frac{7}{10}$. Proved.

From both sides the values head for 12. Or: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$, so for x ≠ 2, $f(x) = x^2 + 2x + 4 \to 4 + 4 + 4 = 12$. Proved.

The graph is the line y = x + 1 with a hole at (1, 2). From both sides it heads for 2:
$\lim_{x \to 1^+} f(x) =$ 2, $\lim_{x \to 1^-} f(x) =$ 2, so $\lim_{x \to 1} f(x) =$ 2.

From both sides the parabola heads for the point (1, 1): 1, 1 and 1.

From both sides the curve heads for (1, 1): 1, 1 and 1.

From both sides the curve heads for the open circle at height 2: 2, 2 and 2. (f(2) itself is not on the graph, but the limit is still 2.)

From the left, the curve comes down to the open circle at height 4: $\lim_{x \to 2^-} f(x) = 4$. From the right, the line starts from the open circle at height 2: $\lim_{x \to 2^+} f(x) = 2$. Since 4 ≠ 2, the limit at x = 2 does not exist.

From the left, the curve rises to the open circle at height 3.5: $\lim_{x \to 1^-} f(x) = 3.5$. From the right, the curve starts from the open circle at height 1: $\lim_{x \to 1^+} f(x) = 1$. The two are not equal, so the limit at x = 1 does not exist. The filled dot, f(1) = 2, doesn’t change that: the limit depends only on the values near 1. (The open circle at (4, 2) is a missing point too, but there both sides head for 2, so $\lim_{x \to 4} f(x) = 2$ does exist.)