Learn4Less Logo

Navigation

Home

Learn

CoursesServicesCompetition MathConcept Review

Tools

WeBWorK WorkshopGrade CalculatorDerivative CalculatorIntegral CalculatorUBC Grade StatisticsMath 100 Readiness Check

More

Free Resources (Blog)ContactJoin Our TeamPrivacy PolicyTerms & Conditions

Our Other Platforms

Limits & Continuity · Topic 03 of 23

Limit Laws and Algebraic Techniques

Try direct substitution first: when ff is continuous at aa, lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a). If substitution gives 00\frac00 the limit may still exist — factor and cancel, multiply by a conjugate or combine fractions until the vanishing factor is gone, then substitute; a bounded oscillation calls for the Squeeze Theorem, and sines for lim⁡x→0sin⁡xx=1\lim_{x\to0}\frac{\sin x}{x}=1.

4 min readCore — on every final7 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Substitute first, when continuity allows

    Polynomials, rational functions (denominator ≠0\neq0), roots, sin⁡\sin, cos⁡\cos, exe^{x} and ln⁡x\ln x are continuous on their domains, so lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a). Try this before any algebra; most limits end here.

  2. 2

    The laws need existing pieces

    Sum, product and quotient laws split a limit only when each piece has a finite limit — and a quotient's denominator limit is not 00. They run one way: lim⁡(f+g)\lim(f+g) existing says nothing about lim⁡f\lim f alone.

  3. 3

    00\frac00 is an instruction, not an answer

    It says: do more algebra. A limit never inspects x=ax=a, so for x≠ax\neq a the factor x−ax-a is a nonzero number you may cancel. Two functions equal away from aa have the same limit at aa.

  4. 4

    Squeeze a bounded oscillation

    If g≤f≤hg\le f\le h near aa and g,h→Lg,h\to L, then f→Lf\to L. Reach for it when a factor like sin⁡1x\sin\frac1x has no limit but is trapped in [−1,1][-1,1]; multiply by something non-negative such as x2x^{2}.

  5. 5

    Two trig limits, radians only

    sin⁡xx→1\frac{\sin x}{x}\to1 and 1−cos⁡xx→0\frac{1-\cos x}{x}\to0 as x→0x\to0. Manufacture them: whatever sits inside the sine must also sit beneath it, so sin⁡5xx=5⋅sin⁡5x5x→5\frac{\sin5x}{x}=5\cdot\frac{\sin5x}{5x}\to5.

-1123-11234xylimit = 2x = 1hole at (1, 2)y = (x² − 1)/(x − 1)
x2−1x−1\frac{x^{2}-1}{x-1} equals x+1x+1 for every x≠1x\neq1, so its graph is that line with one point missing. The limit at 11 never inspects the hole: cancel x−1x-1, substitute, get 22. The function is undefined at 11; the limit is not.
-0.3-0.2-0.10.10.20.3-0.4-0.20.20.4xyy = x·sin(1/x)y = |x|y = −|x|both envelopes → 0, so the trapped curve → 0
sin⁡1x\sin\frac1x swings through [−1,1][-1,1] forever as x→0x\to0, so it has no limit — but ∣xsin⁡1x∣≤∣x∣|x\sin\frac1x|\le|x| traps the curve in the shaded wedge −∣x∣≤xsin⁡1x≤∣x∣-|x|\le x\sin\frac1x\le|x|. Both envelopes (dashed) tend to 00, so the Squeeze Theorem forces the limit 00. The bounds are ±∣x∣\pm|x|, not ±x\pm x: multiplying by xx flips the inequality when x<0x<0.
02

Formulas

What to have memorised

  • Direct substitution

    lim⁡x→af(x)=f(a)if f is continuous at a\lim_{x\to a}f(x)=f(a)\quad\text{if } f \text{ is continuous at } a

    Try it first: a number means done; c0\frac{c}{0} means infinite or no limit; 00\frac00 means algebra.

  • Sum, product and quotient laws

    lim⁡(f±g)=L±M,lim⁡(fg)=LM,lim⁡fg=LM  (M≠0)\lim(f\pm g)=L\pm M,\qquad \lim (fg)=LM,\qquad \lim\frac{f}{g}=\frac{L}{M}\ \ (M\neq0)

    Valid only when L=lim⁡fL=\lim f and M=lim⁡gM=\lim g both exist and are finite.

  • Limit inside a continuous function

    lim⁡x→aF(g(x))=F(lim⁡x→ag(x))\lim_{x\to a}F\bigl(g(x)\bigr)=F\Bigl(\lim_{x\to a}g(x)\Bigr)

    Needs FF continuous at the inner limit — roots, powers, sin⁡\sin, cos⁡\cos, exe^{x}.

  • Conjugates

    (u−v)(u+v)=u−v,(a−b)(a2+ab+b2)=a3−b3(\sqrt{u}-\sqrt{v})(\sqrt{u}+\sqrt{v})=u-v,\qquad (a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}

    Multiply top and bottom by the conjugate; the hidden factor x−ax-a appears.

  • Squeeze Theorem

    g(x)≤f(x)≤h(x) near a,lim⁡x→ag(x)=lim⁡x→ah(x)=L ⟹ lim⁡x→af(x)=Lg(x)\le f(x)\le h(x)\ \text{near } a,\quad \lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L\ \Longrightarrow\ \lim_{x\to a}f(x)=L

    The outer limits must be equal; ff need not be defined at aa.

  • The standard sine limits

    lim⁡x→0sin⁡xx=1,lim⁡x→0sin⁡kxx=k,lim⁡x→0sin⁡axsin⁡bx=ab\lim_{x\to0}\frac{\sin x}{x}=1,\qquad \lim_{x\to0}\frac{\sin kx}{x}=k,\qquad \lim_{x\to0}\frac{\sin ax}{\sin bx}=\frac{a}{b}

    Radians, and only as x→0x\to0. Also tan⁡xx→1\frac{\tan x}{x}\to1.

  • The cosine limits

    lim⁡x→01−cos⁡xx=0,lim⁡x→01−cos⁡xx2=12\lim_{x\to0}\frac{1-\cos x}{x}=0,\qquad \lim_{x\to0}\frac{1-\cos x}{x^{2}}=\frac12

    Multiply by 1+cos⁡x1+cos⁡x\frac{1+\cos x}{1+\cos x}, then use 1−cos⁡2x=sin⁡2x1-\cos^{2}x=\sin^{2}x.

03

Evaluate a limit that gives $\frac00$

The steps, in order
  1. 1

    Substitute x=ax=a. A number: done. c0\frac{c}{0} with c≠0c\neq0: an infinite or nonexistent limit — check the sign on each side. 00\frac00: keep going.

  2. 2

    Polynomials top and bottom: factor both — (x−a)(x-a) divides each — cancel it (legal since x≠ax\neq a), then substitute.

  3. 3

    A square root: multiply top and bottom by its conjugate; for a cube root use a2+ab+b2a^{2}+ab+b^{2}. Then cancel and substitute.

  4. 4

    Small fractions inside a fraction: combine over a common denominator, simplify, cancel, substitute.

  5. 5

    Sines and cosines: put kxkx under every sin⁡kx\sin kx and use sin⁡uu→1\frac{\sin u}{u}\to1; for 1−cos⁡1-\cos multiply by 1+cos⁡1+\cos.

  6. 6

    A bounded oscillating factor: trap the whole function between two bounds with the same limit and quote the Squeeze Theorem.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    00=1\frac00=1 (or =0=0), so the limit is 11

    ✓ Right

    00\frac00 is indeterminate — factor, rationalise or combine fractions, then substitute.

    Why: The symbol records that more algebra is needed; it has no value.

  • ✗ Wrong

    lim⁡x→1x2−1x−1\lim_{x\to1}\frac{x^{2}-1}{x-1} does not exist because you cannot divide by zero

    ✓ Right

    The function is undefined at 11; the limit is lim⁡x→1(x+1)=2\lim_{x\to1}(x+1)=2.

    Why: A limit only looks at xx near aa, never at x=ax=a itself.

  • ✗ Wrong

    lim⁡fg=lim⁡flim⁡g\lim\frac{f}{g}=\frac{\lim f}{\lim g} and lim⁡(f−g)=lim⁡f−lim⁡g\lim(f-g)=\lim f-\lim g, always

    ✓ Right

    Only when both limits exist (and lim⁡g≠0\lim g\neq0 for a quotient). Otherwise simplify first: 1x−1x=0\frac1x-\frac1x=0 has limit 00 at 00 though neither piece has one.

  • ✗ Wrong

    lim⁡x→0sin⁡3xx=1\lim_{x\to0}\frac{\sin3x}{x}=1

    ✓ Right

    =3=3: write 3⋅sin⁡3x3x3\cdot\frac{\sin3x}{3x}. The argument of the sine and the denominator must be identical.

  • ✗ Wrong

    lim⁡x→πsin⁡xx=1\lim_{x\to\pi}\frac{\sin x}{x}=1

    ✓ Right

    The standard limit lives only at x→0x\to0. Here substitution works: sin⁡ππ=0\frac{\sin\pi}{\pi}=0.

    Why: Check where the limit is taken; and in degrees the limit at 00 would be π180\frac{\pi}{180}, not 11.

  • ✗ Wrong

    −x≤xsin⁡1x≤x-x\le x\sin\frac1x\le x, so squeeze to get 00

    ✓ Right

    Multiplying by xx flips the inequality when x<0x<0. Use −∣x∣≤xsin⁡1x≤∣x∣-|x|\le x\sin\frac1x\le|x|, or a non-negative multiplier such as x2x^{2}.

    Why: A squeeze needs genuine bounds on both sides of the point.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Suppose lim⁡x→2f(x)=5\lim_{x\to2}f(x)=5 and lim⁡x→2g(x)=−2\lim_{x\to2}g(x)=-2. Evaluate lim⁡x→22f(x)−g(x)f(x)+g(x)\displaystyle\lim_{x\to2}\frac{2f(x)-g(x)}{f(x)+g(x)}, stating the hypothesis you must check before using the Quotient Law.

    Hint

    Find the limit of the top and of the bottom separately, then look hard at the bottom's value.

    Show answer

    Answer

    44; the denominator's limit lim⁡x→2(f(x)+g(x))=3\lim_{x\to2}\bigl(f(x)+g(x)\bigr)=3 must be nonzero.

    Steps

    Both given limits are finite, so the Constant Multiple, Difference and Sum Laws apply to top and bottom separately:

    lim⁡x→2(2f(x)−g(x))=2(5)−(−2)=12,lim⁡x→2(f(x)+g(x))=5+(−2)=3\lim_{x\to2}\bigl(2f(x)-g(x)\bigr)=2(5)-(-2)=12,\qquad \lim_{x\to2}\bigl(f(x)+g(x)\bigr)=5+(-2)=3

    The Quotient Law needs the denominator's limit to be nonzero. It is 3≠03\neq0, so the limit is 123=4\frac{12}{3}=4. Had it been 00, the law would say nothing — the limit could be finite, infinite or nonexistent.

  2. Q2easy

    Evaluate lim⁡x→−1x2+3x+2x2−1\displaystyle\lim_{x\to-1}\frac{x^{2}+3x+2}{x^{2}-1}.

    Hint

    Substitution gives 00\frac00, so (x+1)(x+1) divides both quadratics.

    Show answer

    Answer

    −12-\dfrac12

    Steps

    At x=−1x=-1 both top and bottom are 00: the form is 00\frac00. Factor and cancel:

    x2+3x+2x2−1=(x+1)(x+2)(x+1)(x−1)=x+2x−1(x≠−1)\frac{x^{2}+3x+2}{x^{2}-1}=\frac{(x+1)(x+2)}{(x+1)(x-1)}=\frac{x+2}{x-1}\qquad(x\neq-1)

    The simplified function is continuous at −1-1 (denominator −2≠0-2\neq0), so substitute: −1+2−1−1=−12\dfrac{-1+2}{-1-1}=-\dfrac12.

  3. Q3medium

    Evaluate lim⁡x→0sin⁡7xsin⁡3x\displaystyle\lim_{x\to0}\frac{\sin 7x}{\sin 3x}.

    Hint

    You know sin⁡uu→1\frac{\sin u}{u}\to1: put a 7x7x under the top sine and a 3x3x under the bottom one without changing the expression.

    Show answer

    Answer

    73\dfrac73

    Steps

    The form is 00\frac00. For x≠0x\neq0 near 00, insert the matching arguments:

    sin⁡7xsin⁡3x=sin⁡7x7x⋅3xsin⁡3x⋅7x3x=73⋅sin⁡7x7x⋅3xsin⁡3x\frac{\sin 7x}{\sin 3x}=\frac{\sin 7x}{7x}\cdot\frac{3x}{\sin 3x}\cdot\frac{7x}{3x}=\frac73\cdot\frac{\sin 7x}{7x}\cdot\frac{3x}{\sin 3x}

    With u=7x→0u=7x\to0 and v=3x→0v=3x\to0, sin⁡uu→1\frac{\sin u}{u}\to1 and vsin⁡v→11=1\frac{v}{\sin v}\to\frac11=1. All three factors have limits, so the Product Law gives 73⋅1⋅1=73\frac73\cdot1\cdot1=\frac73.

  4. Q4medium

    Evaluate lim⁡t→0(1t1+t−1t)\displaystyle\lim_{t\to0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right).

    Hint

    Each term blows up on its own (∞−∞\infty-\infty): combine over one denominator before doing anything else.

    Show answer

    Answer

    −12-\dfrac12

    Steps

    Neither term has a limit at 00, so the Difference Law does not apply. Combine, then rationalise the numerator with 1+1+t1+\sqrt{1+t}:

    1−1+tt1+t⋅1+1+t1+1+t=−tt1+t (1+1+t)=−11+t (1+1+t)(t≠0)\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\cdot\frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}=\frac{-t}{t\sqrt{1+t}\,\bigl(1+\sqrt{1+t}\bigr)}=\frac{-1}{\sqrt{1+t}\,\bigl(1+\sqrt{1+t}\bigr)}\quad(t\neq0)

    The last expression is continuous at t=0t=0: substitute to get −11⋅(1+1)=−12\dfrac{-1}{1\cdot(1+1)}=-\dfrac12.

  5. Q5medium

    Use the Squeeze Theorem to evaluate lim⁡x→0x2esin⁡(1/x)\displaystyle\lim_{x\to0}x^{2}e^{\sin(1/x)}, stating the inequality you use and why it is valid.

    Hint

    Bound the exponent first, then use that ete^{t} is increasing.

    Show answer

    Answer

    00

    Steps

    sin⁡1x\sin\frac1x has no limit at 00, but it is bounded: −1≤sin⁡1x≤1-1\le\sin\frac1x\le1 for every x≠0x\neq0. Since ete^{t} is increasing, e−1≤esin⁡(1/x)≤ee^{-1}\le e^{\sin(1/x)}\le e. Multiplying by x2≥0x^{2}\ge0 keeps the direction:

    x2e≤x2esin⁡(1/x)≤e x2(x≠0)\frac{x^{2}}{e}\le x^{2}e^{\sin(1/x)}\le e\,x^{2}\qquad(x\neq0)

    Both outer functions tend to 00 as x→0x\to0, so by the Squeeze Theorem lim⁡x→0x2esin⁡(1/x)=0\lim_{x\to0}x^{2}e^{\sin(1/x)}=0. The non-negative multiplier x2x^{2} is what keeps the inequality valid on both sides of 00.

  6. Q6hard

    Find constants aa and bb such that lim⁡x→0a+bx−3x=2\displaystyle\lim_{x\to0}\frac{\sqrt{a+bx}-3}{x}=2. Explain why aa is forced to take the value you claim.

    Hint

    If the bottom tends to 0 and the quotient tends to a finite number, what is forced on the top?

    Show answer

    Answer

    a=9a=9, b=12b=12.

    Steps

    Let N(x)=a+bx−3N(x)=\sqrt{a+bx}-3. If N(x)x→2\frac{N(x)}{x}\to2 while x→0x\to0, the Product Law gives N(x)=N(x)x⋅x→2⋅0=0N(x)=\frac{N(x)}{x}\cdot x\to2\cdot0=0, so a−3=0\sqrt a-3=0 and a=9a=9: a nonzero numerator over a vanishing denominator could never converge to 22.

    With a=9a=9, rationalise for x≠0x\neq0:

    9+bx−3x⋅9+bx+39+bx+3=bxx(9+bx+3)=b9+bx+3→b6\frac{\sqrt{9+bx}-3}{x}\cdot\frac{\sqrt{9+bx}+3}{\sqrt{9+bx}+3}=\frac{bx}{x\bigl(\sqrt{9+bx}+3\bigr)}=\frac{b}{\sqrt{9+bx}+3}\to\frac{b}{6}

    Set b6=2\frac b6=2: b=12b=12. Check: 129+12x+3→126=2\frac{12}{\sqrt{9+12x}+3}\to\frac{12}{6}=2. ✓

06

On the exam

How this topic is marked

  • Write the form substitution gives — a number, c0\frac{c}{0} or 00\frac00 — before you do anything. Recognising it earns the method mark; writing 00\frac00 as an answer loses the question.

  • Keep lim⁡x→a\lim_{x\to a} on every line until you substitute, and note "x≠ax\neq a" when you cancel: x2−1x−1\frac{x^{2}-1}{x-1} and x+1x+1 are different functions with the same limit.

  • A squeeze is marked in three parts: the inequality, why it holds near aa (a non-negative multiplier), and that both outer limits are equal. State all three.

Keep going

Stuck on this topic?

One session with a tutor who teaches it every term usually settles it.

Book a session