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Limits & Continuity · Topic 01 of 23

Functions, Graphs and Growth Rates

Before any calculus you must read a function on sight: its natural domain is the intersection of every restriction, and factoring a rational function separates holes from vertical asymptotes. In the long run ln⁡x≪xp≪ax\ln x \ll x^p \ll a^x for every p>0p>0 and a>1a>1, so a limit at infinity is settled by dividing through by the fastest term.

5 min readOccasional on exams8 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Domain: intersect every restriction

    Denominators cannot be 00, even roots need ≥0\ge0 inside, logarithms need >0>0 inside. Impose each condition on its own, then keep only the xx that satisfy all of them.

  2. 2

    Factor before reading a rational function

    A factor that cancels from top and bottom leaves a hole (height from the reduced formula); a zero still in the denominator is a vertical asymptote. Compare degrees for the end behaviour.

  3. 3

    Inside moves xx backwards, outside moves yy

    Factor the argument first: f(2x−6)=f(2(x−3))f(2x-6)=f(2(x-3)) compresses by 12\tfrac12 then shifts right 33, not 66. Outside operations happen in written order: c f(x)+kc\,f(x)+k stretches, then shifts.

  4. 4

    Even, odd, or (usually) neither

    Check the domain is symmetric about 00, then compute f(−x)f(-x) for the whole formula: equal to f(x)f(x) is even, equal to −f(x)-f(x) is odd. One even term does not make a function even.

  5. 5

    Logs, then powers, then exponentials

    As x→∞x\to\infty, (ln⁡x)q/xp→0(\ln x)^q/x^p\to0 and xp/ax→0x^p/a^x\to0 for every p,q>0p,q>0 and a>1a>1. To evaluate a limit at infinity, divide top and bottom by the single fastest term.

0.51.01.52.02.53.03.54.0246810xyy = ln xy = √xy = xy = x²y = eˣ
The growth race: as x→∞x\to\infty every power of ln⁡x\ln x loses to every xpx^p with p>0p>0, and every xpx^p loses to every axa^x with a>1a>1. Within a class, the larger pp or the larger aa wins.
12345610203040xy(2, 4)(4, 16)y = x²y = 2ˣ
x2x^2 is ahead of 2x2^x only on the shaded strip 2<x<42<x<4. After the crossing at (4,16)(4,16) the exponential wins for ever — a growth statement is about x→∞x\to\infty, not about the values you can see.
02

Formulas

What to have memorised

  • Natural-domain checks

    1g(x): g(x)≠0g(x): g(x)≥0ln⁡g(x): g(x)>0\frac{1}{g(x)}:\ g(x)\neq0\qquad \sqrt{g(x)}:\ g(x)\ge0\qquad \ln g(x):\ g(x)>0

    Also arcsin⁡g\arcsin g, arccos⁡g\arccos g need −1≤g≤1-1\le g\le1; tan⁡x\tan x needs x≠π2+kπx\neq\frac{\pi}{2}+k\pi.

  • Horizontal asymptote of P/QP/Q by degrees

    deg⁡P<deg⁡Q ⇒ y=0,deg⁡P=deg⁡Q ⇒ y=lead Plead Q\deg P<\deg Q\ \Rightarrow\ y=0,\qquad \deg P=\deg Q\ \Rightarrow\ y=\frac{\text{lead }P}{\text{lead }Q}

    If deg⁡P=deg⁡Q+1\deg P=\deg Q+1, long-divide: the quotient line is the slant asymptote.

  • Even and odd

    f(−x)=f(x) (even)f(−x)=−f(x) (odd)f(-x)=f(x)\ \text{(even)}\qquad f(-x)=-f(x)\ \text{(odd)}

    Domain must be symmetric about 00. Odd × odd = even; even × odd = odd.

  • Inverse function

    f−1(f(x))=x,f ⁣(f−1(y))=y,dom⁡f−1=ran⁡ff^{-1}(f(x))=x,\qquad f\!\left(f^{-1}(y)\right)=y,\qquad \operatorname{dom}f^{-1}=\operatorname{ran}f

    Exists exactly when ff is one-to-one (horizontal line test). Graph: reflect in y=xy=x.

  • Every exponential is a rescaled exe^x

    ax=exln⁡a,log⁡ax=ln⁡xln⁡aa^x=e^{x\ln a},\qquad \log_a x=\frac{\ln x}{\ln a}

    To compare 2x2^x with ex/2e^{x/2}, compare exponents: ln⁡2≈0.693>0.5\ln 2\approx0.693>0.5.

  • Logarithm laws (u,v>0u,v>0)

    ln⁡(uv)=ln⁡u+ln⁡v,ln⁡uv=ln⁡u−ln⁡v,ln⁡ur=rln⁡u\ln(uv)=\ln u+\ln v,\qquad \ln\frac{u}{v}=\ln u-\ln v,\qquad \ln u^{r}=r\ln u

    ln⁡(u+v)\ln(u+v) does not split. ln⁡x2=2ln⁡∣x∣\ln x^2=2\ln|x|, valid for all x≠0x\neq0.

  • Growth hierarchy

    lim⁡x→∞(ln⁡x)qxp=0,lim⁡x→∞xpax=0,lim⁡x→0+xpln⁡x=0\lim_{x\to\infty}\frac{(\ln x)^{q}}{x^{p}}=0,\qquad \lim_{x\to\infty}\frac{x^{p}}{a^{x}}=0,\qquad \lim_{x\to0^{+}}x^{p}\ln x=0

    For every p,q>0p,q>0 and a>1a>1. Larger power or larger base wins.

  • Sinusoid y=Asin⁡(B(x−C))+Dy=A\sin(B(x-C))+D

    amplitude ∣A∣,period 2π∣B∣,shift C,midline y=D\text{amplitude }|A|,\quad \text{period }\frac{2\pi}{|B|},\quad \text{shift }C,\quad \text{midline }y=D

    Factor BB out of the argument before reading CC. Range [D−∣A∣, D+∣A∣][D-|A|,\,D+|A|].

03

Read a rational function without calculus

The steps, in order
  1. 1

    Factor numerator and denominator completely. The domain excludes every zero of the original denominator.

  2. 2

    Cancel common factors. Each cancelled zero is a hole; its height is the reduced formula evaluated there.

  3. 3

    Each remaining zero of the denominator is a vertical asymptote. Read the sign on each side from the reduced form.

  4. 4

    Compare degrees: smaller on top gives y=0y=0; equal gives the ratio of leading coefficients; one higher on top — long-divide for the slant asymptote.

  5. 5

    Find the intercepts (f(0)f(0) and the zeros of the reduced numerator), then check whether f(x)=f(x)= asymptote has a solution.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    x2=x\sqrt{x^2}=x, so x2−9\sqrt{x^2-9} needs x≥3x\ge3

    ✓ Right

    x2=∣x∣\sqrt{x^2}=|x|; x2≥9x^2\ge9 means ∣x∣≥3|x|\ge3, so the domain is (−∞,−3]∪[3,∞)(-\infty,-3]\cup[3,\infty).

    Why: The square root symbol always returns the non-negative root.

  • ✗ Wrong

    ln⁡(a+b)=ln⁡a+ln⁡b\ln(a+b)=\ln a+\ln b

    ✓ Right

    Only products split: ln⁡(ab)=ln⁡a+ln⁡b\ln(ab)=\ln a+\ln b for a,b>0a,b>0. ln⁡(a+b)\ln(a+b) does not simplify.

  • ✗ Wrong

    y=f(2x−6)y=f(2x-6): compress horizontally by 12\tfrac12, then shift right 66

    ✓ Right

    Factor first: f(2(x−3))f(2(x-3)) — compress by 12\tfrac12, then shift right 33.

    Why: Compressing, then shifting by 66, produces f(2(x−6))=f(2x−12)f(2(x-6))=f(2x-12) — a different graph.

  • ✗ Wrong

    f(x)=x2+xf(x)=x^2+x is even because x2x^2 is

    ✓ Right

    f(−1)=0f(-1)=0 but f(1)=2f(1)=2: neither even nor odd.

    Why: One term never decides; compute f(−x)f(-x) for the whole formula.

  • ✗ Wrong

    x2−4x2−x−6=x−2x−3\dfrac{x^2-4}{x^2-x-6}=\dfrac{x-2}{x-3}, so nothing happens at x=−2x=-2

    ✓ Right

    Equal only for x≠−2x\neq-2; the graph has a hole at (−2,45)\left(-2,\tfrac45\right), not a point.

    Why: Cancelling changes the formula, not the domain.

  • ✗ Wrong

    x100x^{100} grows faster than 2x2^x — just check x=10x=10

    ✓ Right

    2x2^x is larger for every xx beyond about 996996: growth statements are about x→∞x\to\infty only.

    Why: xp/ax→0x^{p}/a^{x}\to0 for every pp and every a>1a>1.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Find the natural domain of f(x)=x+3x2−4f(x)=\dfrac{\sqrt{x+3}}{x^2-4} in interval notation.

    Hint

    Two separate requirements — impose both, then intersect.

    Show answer

    Answer

    [−3,−2)∪(−2,2)∪(2,∞)[-3,-2)\cup(-2,2)\cup(2,\infty)

    Steps

    Root: x+3≥0x+3\ge0, so x≥−3x\ge-3. Denominator: x2−4=(x−2)(x+2)≠0x^2-4=(x-2)(x+2)\neq0, so x≠±2x\neq\pm2.

    Both −2-2 and 22 lie inside [−3,∞)[-3,\infty), so delete them: [−3,−2)∪(−2,2)∪(2,∞)[-3,-2)\cup(-2,2)\cup(2,\infty)

    Check the endpoint: f(−3)=05=0f(-3)=\frac{0}{5}=0 is defined, so −3-3 is included.

  2. Q2easy

    Classify each function as even, odd or neither, with justification.

    (a) x4−3x2+7x^4-3x^2+7 (b) x3sin⁡xx^3\sin x (c) xx2+1\dfrac{x}{x^2+1} (d) ln⁡x\ln x

    Hint

    Check that the domain is symmetric about 00 first, then compute f(−x)f(-x).

    Show answer

    Answer

    (a) even (b) even (c) odd (d) neither

    Steps

    (a) (−x)4−3(−x)2+7=x4−3x2+7(-x)^4-3(-x)^2+7=x^4-3x^2+7: even.

    (b) (−x)3sin⁡(−x)=(−x3)(−sin⁡x)=x3sin⁡x(-x)^3\sin(-x)=(-x^3)(-\sin x)=x^3\sin x: even — odd times odd is even.

    (c) −x(−x)2+1=−xx2+1\dfrac{-x}{(-x)^2+1}=-\dfrac{x}{x^2+1}: odd.

    (d) The domain (0,∞)(0,\infty) is not symmetric about 00, so f(−x)f(-x) is not even defined: neither.

  3. Q3medium

    For f(x)=2x2−3x−2x2−4f(x)=\dfrac{2x^2-3x-2}{x^2-4} find the domain, any holes, all asymptotes and both intercepts, and describe the graph on each side of the vertical asymptote.

    Hint

    Factor top and bottom first; a common factor changes the story.

    Show answer

    Answer

    Domain x≠±2x\neq\pm2; hole at (2,54)\left(2,\tfrac54\right); vertical asymptote x=−2x=-2, horizontal asymptote y=2y=2; intercepts (−12,0)\left(-\tfrac12,0\right) and (0,12)\left(0,\tfrac12\right); f→+∞f\to+\infty as x→−2−x\to-2^- and f→−∞f\to-\infty as x→−2+x\to-2^+.

    Steps

    f(x)=(2x+1)(x−2)(x−2)(x+2)=2x+1x+2(x≠2)f(x)=\frac{(2x+1)(x-2)}{(x-2)(x+2)}=\frac{2x+1}{x+2}\quad(x\neq2)

    Domain: x≠±2x\neq\pm2. The cancelled factor gives a hole at x=2x=2 of height 54\frac{5}{4}. The reduced denominator vanishes only at x=−2x=-2: vertical asymptote. Equal degrees with leading coefficients 22 and 11: horizontal asymptote y=2y=2.

    Intercepts: 2x+1=02x+1=0 gives x=−12x=-\frac12; f(0)=−2−4=12f(0)=\frac{-2}{-4}=\frac12.

    Near x=−2x=-2 the numerator is about −3-3. As x→−2−x\to-2^-, x+2→0−x+2\to0^- so f→+∞f\to+\infty; as x→−2+x\to-2^+, x+2→0+x+2\to0^+ so f→−∞f\to-\infty.

  4. Q4medium

    Solve log⁡2x+log⁡2(x−2)=3\log_2 x+\log_2(x-2)=3, stating which candidate solutions are valid and why.

    Hint

    Write down the domain before you combine the logarithms.

    Show answer

    Answer

    x=4x=4 only; x=−2x=-2 is extraneous.

    Steps

    Domain: x>0x>0 and x−2>0x-2>0, so x>2x>2.

    For such xx the product law applies: log⁡2(x(x−2))=3\log_2\big(x(x-2)\big)=3, so x(x−2)=23=8x(x-2)=2^3=8. x2−2x−8=(x−4)(x+2)=0x^2-2x-8=(x-4)(x+2)=0

    x=−2x=-2 fails x>2x>2 (indeed log⁡2(−2)\log_2(-2) is undefined): reject. x=4x=4: log⁡24+log⁡22=2+1=3\log_2 4+\log_2 2=2+1=3. ✓

  5. Q5medium

    Show that f(x)=2x+1x−3f(x)=\dfrac{2x+1}{x-3} is one-to-one, find a formula for f−1(x)f^{-1}(x), and state the domain and range of ff and of f−1f^{-1}.

    Hint

    Write f(x)f(x) as a constant plus a multiple of 1x−3\frac{1}{x-3}.

    Show answer

    Answer

    f−1(x)=3x+1x−2f^{-1}(x)=\dfrac{3x+1}{x-2}. dom⁡f={x≠3}=ran⁡f−1\operatorname{dom}f=\{x\neq3\}=\operatorname{ran}f^{-1} and ran⁡f={y≠2}=dom⁡f−1\operatorname{ran}f=\{y\neq2\}=\operatorname{dom}f^{-1}.

    Steps

    f(x)=2(x−3)+7x−3=2+7x−3f(x)=\dfrac{2(x-3)+7}{x-3}=2+\dfrac{7}{x-3}. If f(a)=f(b)f(a)=f(b) then 7a−3=7b−3\frac{7}{a-3}=\frac{7}{b-3}, so a=ba=b: one-to-one. Since 7x−3\frac{7}{x-3} takes every value except 00, the range of ff is y≠2y\neq2.

    Solve y=2x+1x−3y=\dfrac{2x+1}{x-3} for xx: yx−3y=2x+1yx-3y=2x+1, so x(y−2)=3y+1x(y-2)=3y+1 and f−1(x)=3x+1x−2f^{-1}(x)=\frac{3x+1}{x-2}

    Domain and range swap: dom⁡f−1={x≠2}\operatorname{dom}f^{-1}=\{x\neq2\}, ran⁡f−1={y≠3}\operatorname{ran}f^{-1}=\{y\neq3\}. Check: f(4)=9f(4)=9 and f−1(9)=287=4f^{-1}(9)=\frac{28}{7}=4.

  6. Q6hard

    Evaluate without l'Hôpital's rule, justifying each step with the growth hierarchy.

    (a) lim⁡x→∞3x+x102⋅3x+5x/2\displaystyle\lim_{x\to\infty}\frac{3^x+x^{10}}{2\cdot3^x+5^{x/2}} (b) lim⁡x→0+xln⁡x\displaystyle\lim_{x\to0^+}x\ln x

    Hint

    (a) Name the fastest term and divide by it. (b) Substitute t=1/xt=1/x.

    Show answer

    Answer

    (a) 12\tfrac12 (b) 00

    Steps

    (a) 5x/2=(5)x5^{x/2}=(\sqrt{5})^x with 5≈2.24<3\sqrt{5}\approx2.24<3, so 3x3^x is the fastest term. Divide top and bottom by it: 1+x10/3x2+(5/3)x→1+02+0=12\frac{1+x^{10}/3^x}{2+(\sqrt{5}/3)^x}\to\frac{1+0}{2+0}=\frac12 since x10/3x→0x^{10}/3^x\to0 (power against exponential) and (5/3)x→0(\sqrt{5}/3)^x\to0 (base below 11).

    (b) Put t=1/xt=1/x, so t→∞t\to\infty: xln⁡x=1tln⁡1t=−ln⁡tt→0x\ln x=\frac1t\ln\frac1t=-\frac{\ln t}{t}\to0 (log against power). The limit is 00, approached from below.

06

On the exam

How this topic is marked

  • Domain and asymptote questions are marked on the factored form: show the factoring, name each cancelled factor as a hole, and give asymptotes as equations (x=−2x=-2, y=2y=2), not numbers.

  • For even/odd and inverse questions, write the domain check first — a missing symmetric-domain or one-to-one line costs the justification mark.

  • Growth-rate limits want the dominant term named and divided out; a table of values or the word "obviously" earns nothing.

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