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Limits & Continuity · Topic 02 of 23

Introduction to Limits

lim⁡x→af(x)=L\lim_{x\to a}f(x)=L says f(x)f(x) gets as close to LL as demanded once xx is close enough to aa — with x≠ax\neq a, so f(a)f(a) never matters. The limit exists exactly when both one-sided limits exist and agree; a jump, a blow-up to ±∞\pm\infty or an oscillation means it does not, and the ε\varepsilon-δ\delta definition makes all of this precise.

5 min readCore — on every final7 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    The limit never looks at x=ax=a

    lim⁡x→af(x)=L\lim_{x\to a}f(x)=L means f(x)f(x) stays as close to LL as demanded once xx is close enough to aa, with x≠ax\neq a. Whether f(a)f(a) equals LL, differs from it, or is undefined changes nothing.

  2. 2

    Both one-sided limits must agree

    lim⁡x→a−\lim_{x\to a^-} uses only x<ax<a; lim⁡x→a+\lim_{x\to a^+} only x>ax>a. lim⁡x→af(x)=L\lim_{x\to a}f(x)=L exactly when both equal LL. At a domain endpoint (x\sqrt{x} at 00) only the one-sided limit is posed.

  3. 3

    Three ways a limit fails

    A jump (one-sided limits differ), a blow-up (1x2\frac{1}{x^2} or 1x\frac1x at 00: a vertical asymptote), or an oscillation (sin⁡1x\sin\frac1x at 00 sweeps [−1,1][-1,1] forever). Writing =∞=\infty records how it fails; ∞\infty is not a number.

  4. 4

    Circles give limits, dots give values

    Slide a finger along the curve from the left, then from the right, and report the height each branch heads toward — an open circle still counts. A solid dot gives f(a)f(a) only; never jump to it.

  5. 5

    ε\varepsilon-δ\delta: every tolerance must be met

    For every ε>0\varepsilon>0 there must be a δ>0\delta>0 with 0<∣x−a∣<δ⇒∣f(x)−L∣<ε0<|x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon. δ\delta may depend on ε\varepsilon and aa, never on xx. One δ\delta for one ε\varepsilon proves nothing.

-2-1123412345xyL = 2hole: the limit is 2f(2) = 3.5y = f(x)
f(x)=x24+1f(x)=\frac{x^2}{4}+1 for x≠2x\neq2, but f(2)=3.5f(2)=3.5. Both branches head to height 22, so lim⁡x→2f(x)=2\lim_{x\to2}f(x)=2. The solid dot is f(2)f(2) and plays no part in the limit — cover it with your thumb and nothing changes.
12341234xyx = 2left limit 2f(2) = 3 = right limity = x²/2 for x < 2y = 5 − x for x ≥ 2
A jump at x=2x=2: lim⁡x→2−f(x)=2\lim_{x\to2^-}f(x)=2 and lim⁡x→2+f(x)=3\lim_{x\to2^+}f(x)=3. Both one-sided limits exist but disagree, so lim⁡x→2f(x)\lim_{x\to2}f(x) does not exist. The solid dot says f(2)=3f(2)=3 — a third fact, independent of either limit.
02

Formulas

What to have memorised

  • Two-sided from one-sided

    lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=L=lim⁡x→a+f(x)\lim_{x\to a}f(x)=L\iff\lim_{x\to a^-}f(x)=L=\lim_{x\to a^+}f(x)

    Both must exist and agree; either failure kills the two-sided limit. At a domain endpoint only one side is posed.

  • ε\varepsilon-δ\delta definition of lim⁡x→af(x)=L\lim_{x\to a}f(x)=L

    ∀ ε>0  ∃ δ>0:0<∣x−a∣<δ ⟹ ∣f(x)−L∣<ε\forall\,\varepsilon>0\ \ \exists\,\delta>0:\quad 0<|x-a|<\delta\ \Longrightarrow\ |f(x)-L|<\varepsilon

    ε\varepsilon is the opponent's vertical tolerance; δ\delta is your horizontal reply, written in terms of ε\varepsilon only.

  • One-sided limit, precisely

    lim⁡x→a+f(x)=L  ⟺  ∀ ε>0  ∃ δ>0: a<x<a+δ ⟹ ∣f(x)−L∣<ε\lim_{x\to a^+}f(x)=L\iff\forall\,\varepsilon>0\ \ \exists\,\delta>0:\ a<x<a+\delta\ \Longrightarrow\ |f(x)-L|<\varepsilon

    Left-hand: use a−δ<x<aa-\delta<x<a instead. The superscript names the side xx comes from, not the sign of ff.

  • δ\delta for a linear function

    δ=ε∣m∣for f(x)=mx+b, m≠0\delta=\frac{\varepsilon}{|m|}\quad\text{for } f(x)=mx+b,\ m\neq0

    Because ∣f(x)−(ma+b)∣=∣m∣ ∣x−a∣|f(x)-(ma+b)|=|m|\,|x-a|. If m=0m=0 the function is constant and any δ>0\delta>0 works.

  • δ\delta with a nuisance factor

    δ=min⁡{1, εK}where ∣x−a∣<1⇒∣g(x)∣<K\delta=\min\left\{1,\ \frac{\varepsilon}{K}\right\}\quad\text{where } |x-a|<1\Rightarrow|g(x)|<K

    For ∣f(x)−L∣=∣x−a∣ ∣g(x)∣|f(x)-L|=|x-a|\,|g(x)|: restrict first, bound the other factor, then take the minimum.

  • Infinite limit

    lim⁡x→af(x)=∞  ⟺  ∀ M>0  ∃ δ>0: 0<∣x−a∣<δ⟹f(x)>M\lim_{x\to a}f(x)=\infty\iff\forall\,M>0\ \ \exists\,\delta>0:\ 0<|x-a|<\delta\Longrightarrow f(x)>M

    A description of failure — the limit does not exist, because ∞\infty is not a real number.

  • Showing a limit does not exist

    xn→a,  tn→a,∣f(xn)−f(tn)∣≥c>0 ⟹ lim⁡x→af(x) does not existx_n\to a,\ \ t_n\to a,\quad |f(x_n)-f(t_n)|\ge c>0\ \Longrightarrow\ \lim_{x\to a}f(x)\ \text{does not exist}

    Two families of inputs whose outputs stay apart: for sin⁡1x\sin\frac1x take xn=12πnx_n=\frac{1}{2\pi n}, tn=12πn+π/2t_n=\frac{1}{2\pi n+\pi/2}.

03

Write an $\varepsilon$-$\delta$ proof

The steps, in order
  1. 1

    Scratch work first: simplify ∣f(x)−L∣|f(x)-L| until ∣x−a∣|x-a| appears as a factor — ∣m∣ ∣x−a∣|m|\,|x-a| for a line, ∣x−a∣ ∣x+a∣|x-a|\,|x+a| for x2x^2.

  2. 2

    If the other factor still contains xx, agree δ≤1\delta\le1: then ∣x−a∣<1|x-a|<1 pins xx to an interval and bounds that factor by a constant KK.

  3. 3

    Choose δ\delta: ε/∣m∣\varepsilon/|m| for a line, min⁡{1,ε/K}\min\{1,\varepsilon/K\} otherwise. It may involve ε\varepsilon and aa, never xx.

  4. 4

    Write the proof: "Let ε>0\varepsilon>0. Put δ=…\delta=\dots. Suppose 0<∣x−a∣<δ0<|x-a|<\delta. Then ∣f(x)−L∣=⋯<⋯=ε|f(x)-L|=\dots<\dots=\varepsilon."

  5. 5

    Close with "since ε>0\varepsilon>0 was arbitrary". For lim⁡=∞\lim=\infty, swap in: given M>0M>0, find δ\delta forcing f(x)>Mf(x)>M.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a), automatically

    ✓ Right

    The limit never inspects f(a)f(a). That equation is continuity at aa — a property to verify, not part of the definition.

    Why: f(a)f(a) may differ from the limit or not exist at all.

  • ✗ Wrong

    "lim⁡x→01x2=∞\lim_{x\to0}\frac{1}{x^2}=\infty, so the limit exists."

    ✓ Right

    It does not exist; "=∞=\infty" only records how it fails.

    Why: ∞\infty is not a real number, and existence needs a real LL.

  • ✗ Wrong

    δ=ε∣x+3∣\delta=\dfrac{\varepsilon}{|x+3|}

    ✓ Right

    δ\delta may not mention xx. Restrict δ≤1\delta\le1, bound ∣x+3∣<7|x+3|<7, then take δ=min⁡{1,ε/7}\delta=\min\{1,\varepsilon/7\}.

    Why: xx is quantified after δ\delta; δ\delta must be fixed before xx is chosen.

  • ✗ Wrong

    "The right-hand limit is 44, so lim⁡x→af(x)=4\lim_{x\to a}f(x)=4."

    ✓ Right

    Both one-sided limits must exist and agree — unless aa is a domain endpoint, where only one side is posed.

  • ✗ Wrong

    sin⁡πx\sin\frac{\pi}{x} at x=0.1, 0.01, 0.001x=0.1,\,0.01,\,0.001 gives 0,0,00,0,0, so the limit is 00

    ✓ Right

    Those samples all sit on zeros of sine; lim⁡x→0sin⁡πx\lim_{x\to0}\sin\frac{\pi}{x} does not exist.

    Why: A table samples finitely many points; a limit is a claim about all of them.

  • ✗ Wrong

    "For ε=0.06\varepsilon=0.06, δ=0.02\delta=0.02 works. QED."

    ✓ Right

    Give δ\delta as a formula in ε\varepsilon (δ=ε/3\delta=\varepsilon/3) and run the argument for an arbitrary ε>0\varepsilon>0.

    Why: The definition is a promise about every tolerance, not one convenient one.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Let f(x)={x+1,x<25,x=25−x,x>2f(x)=\begin{cases}x+1, & x<2\\ 5, & x=2\\ 5-x, & x>2\end{cases} Find (a) lim⁡x→2−f(x)\lim_{x\to2^-}f(x), (b) lim⁡x→2+f(x)\lim_{x\to2^+}f(x), (c) lim⁡x→2f(x)\lim_{x\to2}f(x), (d) f(2)f(2). What is the relationship between (c) and (d)?

    Hint

    Each one-sided limit uses only the formula valid on its side of 22; the value at 22 is a separate question.

    Show answer

    Answer

    (a) 33 (b) 33 (c) 33 (d) f(2)=5f(2)=5. The limit exists but differs from the value: a removable discontinuity.

    Steps

    (a) For x<2x<2 the rule is x+1x+1: lim⁡x→2−f(x)=2+1=3\lim_{x\to2^-}f(x)=2+1=3.

    (b) For x>2x>2 the rule is 5−x5-x: lim⁡x→2+f(x)=5−2=3\lim_{x\to2^+}f(x)=5-2=3.

    (c) Both one-sided limits exist and equal 33, so lim⁡x→2f(x)=3\lim_{x\to2}f(x)=3.

    (d) The middle line applies only at x=2x=2: f(2)=5f(2)=5.

    The limit is computed from x≠2x\neq2 and never sees f(2)f(2), so 3≠53\neq5 is perfectly legal. Redefining f(2)=3f(2)=3 would make ff continuous there — that is why the discontinuity is called removable.

  2. Q2easy

    The graph of hh on [−3,4][-3,4] has three pieces: a segment from the solid point (−3,−1)(-3,-1) up to an open circle at (1,3)(1,3); a single solid dot at (1,0)(1,0); a segment from an open circle at (1,1)(1,1) up to the solid point (4,4)(4,4). Find lim⁡x→1−h(x)\lim_{x\to1^-}h(x), lim⁡x→1+h(x)\lim_{x\to1^+}h(x), lim⁡x→1h(x)\lim_{x\to1}h(x), h(1)h(1) and lim⁡x→4−h(x)\lim_{x\to4^-}h(x). Why is lim⁡x→4h(x)\lim_{x\to4}h(x) not asked?

    Hint

    An open circle still tells you the height a branch is heading toward; only the solid dot gives a value.

    Show answer

    Answer

    33; 11; does not exist (jump); h(1)=0h(1)=0; 44. Nothing is defined for x>4x>4, so only the one-sided limit is meaningful there.

    Steps

    Left piece: through (−3,−1)(-3,-1) and (1,3)(1,3), slope 11, so h(x)=x+2h(x)=x+2 for −3≤x<1-3\le x<1. Right piece: through (1,1)(1,1) and (4,4)(4,4), so h(x)=xh(x)=x for 1<x≤41<x\le4.

    lim⁡x→1−h(x)=1+2=3\lim_{x\to1^-}h(x)=1+2=3 — the circle at (1,3)(1,3) being open is irrelevant. lim⁡x→1+h(x)=1\lim_{x\to1^+}h(x)=1. The one-sided limits differ, so lim⁡x→1h(x)\lim_{x\to1}h(x) does not exist. h(1)=0h(1)=0 from the solid dot: left limit, right limit and value are three independent numbers.

    lim⁡x→4−h(x)=4\lim_{x\to4^-}h(x)=4. A two-sided limit needs hh defined on both sides of 44; there is no graph for x>4x>4, so only the left-hand limit is posed.

  3. Q3medium

    Let f(x)=cx+3f(x)=cx+3 for x<1x<1 and f(x)=x2−cxf(x)=x^2-cx for x≥1x\ge1, where cc is a constant. (a) Find every cc for which lim⁡x→1f(x)\lim_{x\to1}f(x) exists, and give the limit. (b) For that cc, compare f(1)f(1) with the limit. (c) If c=0c=0, find both one-sided limits and describe how the limit fails.

    Hint

    Write each one-sided limit as an expression in cc, then force them to agree.

    Show answer

    Answer

    (a) c=−1c=-1 only; the limit is 22. (b) f(1)=2f(1)=2, equal to the limit. (c) Left limit 33, right limit 11: a jump of size 22.

    Steps

    (a) lim⁡x→1−f(x)=c+3\lim_{x\to1^-}f(x)=c+3 and lim⁡x→1+f(x)=1−c\lim_{x\to1^+}f(x)=1-c. The two-sided limit exists exactly when c+3=1−c⟹c=−1,c+3=1-c\quad\Longrightarrow\quad c=-1, and then both sides equal 22.

    (b) The second rule covers x=1x=1: f(1)=1−(−1)(1)=2f(1)=1-(-1)(1)=2. Limit equals value, so ff is continuous at 11 when c=−1c=-1.

    (c) With c=0c=0: f(x)=3f(x)=3 for x<1x<1 and f(x)=x2f(x)=x^2 for x≥1x\ge1, so the one-sided limits are 33 and 11. Both exist but disagree — a jump, not a blow-up or an oscillation.

  4. Q4medium

    Let F(x)=1x−2F(x)=\dfrac{1}{x-2} and G(x)=1(x−2)2G(x)=\dfrac{1}{(x-2)^2}. (a) Find lim⁡x→2−F(x)\lim_{x\to2^-}F(x) and lim⁡x→2+F(x)\lim_{x\to2^+}F(x); what does this say about lim⁡x→2F(x)\lim_{x\to2}F(x)? (b) Prove from the definition that lim⁡x→2G(x)=∞\lim_{x\to2}G(x)=\infty. (c) Does lim⁡x→2G(x)\lim_{x\to2}G(x) exist?

    Hint

    For (a) track the sign of x−2x-2 on each side. For (b) solve G(x)>MG(x)>M for ∣x−2∣|x-2|.

    Show answer

    Answer

    (a) −∞-\infty and +∞+\infty; lim⁡x→2F(x)\lim_{x\to2}F(x) does not exist (vertical asymptote). (b) δ=1/M\delta=1/\sqrt{M} works. (c) No — "=∞=\infty" describes the failure; ∞\infty is not a real number.

    Steps

    (a) As x→2−x\to2^-, x−2x-2 is negative and tiny, so F(x)→−∞F(x)\to-\infty; as x→2+x\to2^+ it is positive and tiny, so F(x)→+∞F(x)\to+\infty. Neither one-sided limit is a real number, and they run to opposite infinities, so lim⁡x→2F(x)\lim_{x\to2}F(x) does not exist; x=2x=2 is a vertical asymptote.

    (b) Let M>0M>0 and put δ=1M\delta=\dfrac{1}{\sqrt{M}}. If 0<∣x−2∣<δ0<|x-2|<\delta then 0<(x−2)2<δ2=1M⟹G(x)=1(x−2)2>M.0<(x-2)^2<\delta^2=\frac{1}{M}\quad\Longrightarrow\quad G(x)=\frac{1}{(x-2)^2}>M. Since M>0M>0 was arbitrary, lim⁡x→2G(x)=∞\lim_{x\to2}G(x)=\infty.

    (c) No. Existence requires a real LL; the symbol ∞\infty records that GG exceeds every bound near 22.

  5. Q5medium

    Let f(x)=3x+2f(x)=3x+2. (a) Find the largest δ>0\delta>0 such that 0<∣x−1∣<δ0<|x-1|<\delta implies ∣f(x)−5∣<0.06|f(x)-5|<0.06. (b) Repeat for ε=0.003\varepsilon=0.003. (c) Give δ\delta as a formula in an arbitrary ε>0\varepsilon>0 and write the full proof that lim⁡x→1(3x+2)=5\lim_{x\to1}(3x+2)=5.

    Hint

    Simplify ∣f(x)−5∣|f(x)-5| first; for a linear function it is a constant times ∣x−1∣|x-1|.

    Show answer

    Answer

    (a) δ=0.02\delta=0.02 (b) δ=0.001\delta=0.001 (c) δ=ε/3\delta=\varepsilon/3.

    Steps

    ∣f(x)−5∣=∣3x−3∣=3∣x−1∣|f(x)-5|=|3x-3|=3|x-1|.

    (a) 3∣x−1∣<0.06  ⟺  ∣x−1∣<0.023|x-1|<0.06\iff|x-1|<0.02, so the largest δ\delta is 0.020.02 (any smaller positive δ\delta also works). (b) ∣x−1∣<0.001|x-1|<0.001: δ=0.001\delta=0.001.

    (c) Let ε>0\varepsilon>0 be given. Put δ=ε/3>0\delta=\varepsilon/3>0. Suppose 0<∣x−1∣<δ0<|x-1|<\delta. Then ∣(3x+2)−5∣=3∣x−1∣<3δ=ε.|(3x+2)-5|=3|x-1|<3\delta=\varepsilon. Since ε>0\varepsilon>0 was arbitrary, lim⁡x→1(3x+2)=5\lim_{x\to1}(3x+2)=5. Check: ε=0.06\varepsilon=0.06 gives δ=0.02\delta=0.02, matching (a).

  6. Q6hard

    Prove from the ε\varepsilon-δ\delta definition that lim⁡x→2x2=4\lim_{x\to2}x^2=4. Say exactly where the preliminary restriction δ≤1\delta\le1 is used and why it is needed.

    Hint

    Factor x2−4x^2-4. You control ∣x−2∣|x-2| directly; the factor ∣x+2∣|x+2| must be bounded first.

    Show answer

    Answer

    δ=min⁡{1,ε/5}\delta=\min\{1,\varepsilon/5\}. The restriction δ≤1\delta\le1 is used only to get ∣x+2∣<5|x+2|<5; without it ∣x+2∣|x+2| could be arbitrarily large.

    Steps

    Scratch: ∣x2−4∣=∣x−2∣ ∣x+2∣|x^2-4|=|x-2|\,|x+2|. If ∣x−2∣<1|x-2|<1 then 1<x<31<x<3, so ∣x+2∣<5|x+2|<5 and ∣x2−4∣<5∣x−2∣|x^2-4|<5|x-2|.

    Proof. Let ε>0\varepsilon>0. Put δ=min⁡{1,ε/5}>0\delta=\min\{1,\varepsilon/5\}>0 and suppose 0<∣x−2∣<δ0<|x-2|<\delta. Since δ≤1\delta\le1, ∣x+2∣=∣(x−2)+4∣≤∣x−2∣+4<5|x+2|=|(x-2)+4|\le|x-2|+4<5. Since δ≤ε/5\delta\le\varepsilon/5, ∣x−2∣<ε/5|x-2|<\varepsilon/5. Hence ∣x2−4∣=∣x−2∣ ∣x+2∣<ε5⋅5=ε.|x^2-4|=|x-2|\,|x+2|<\frac{\varepsilon}{5}\cdot5=\varepsilon. Since ε>0\varepsilon>0 was arbitrary, lim⁡x→2x2=4\lim_{x\to2}x^2=4.

    The restriction is used only for the bound ∣x+2∣<5|x+2|<5. Without it xx could be far from 22, making ∣x+2∣|x+2| huge and the product large even when ∣x−2∣|x-2| is small. Any restriction works: δ≤12\delta\le\tfrac12 gives ∣x+2∣<4.5|x+2|<4.5 and δ=min⁡{12,2ε9}\delta=\min\{\tfrac12,\tfrac{2\varepsilon}{9}\}.

06

On the exam

How this topic is marked

  • "Find the limit or explain why it does not exist" is marked on the reason: write both one-sided values, then "they differ, so the limit does not exist" — or name the blow-up or oscillation.

  • An ε\varepsilon-δ\delta proof is marked line by line: "Let ε>0\varepsilon>0", δ\delta as a formula in ε\varepsilon, "suppose 0<∣x−a∣<δ0<|x-a|<\delta", then a chain of inequalities ending in <ε<\varepsilon. Scratch work alone earns nothing.

  • Never write lim⁡x→af(x)=f(a)\lim_{x\to a}f(x)=f(a) without a reason. Here the reason is "both one-sided limits equal f(a)f(a)"; later it will be continuity.

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