Worked examples — Continuity — definition, types of discontinuity (removable, jump, infinite)

Everything here rests on the parent Continuity note and on Limits — definition and one-sided limits. If a word below feels unfamiliar, it is defined the first time it appears.

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The scenario matrix

Read the table as: "Here is a shape the world can hand you, here is which of the three VLE conditions breaks, and here is the example that drills it." (VLE = Value, Limit, Equal — the mnemonic from the parent.)

#	Case class	What arrives	Which condition breaks	Verdict expected	Example
A	$0/0$ that cancels	rational, factor kills the zero	V fails (undefined)	removable	Ex 1
B	$0/0$ that does NOT cancel	rational, denom stronger	V + L fail	infinite	Ex 2
C	Piecewise, pieces agree	two formulas meet cleanly	none — continuous!	continuous	Ex 3
D	Piecewise, pieces disagree	stair-step	E fails, $L^{-}\ne L^{+}$	jump	Ex 4
E	Wall with matching signs	$1/(x-a{)}^2$ type	V + L fail (both $+\infty$)	infinite (same sign)	Ex 5
F	Oscillation — no settle	$\sin (1/x)$ type	L fails (no limit)	essential (not removable)	Ex 6
G	Real-world word problem	parking-cost step function	E fails at breakpoints	jump	Ex 7
H	Exam twist — find $k$	piecewise with unknown constant	force E to hold	solve for $k$	Ex 8

The columns "signs/quadrants" become left-vs-right sign of the denominator (cases B, E), "zero/degenerate input" is the $0/0$ column (A, B), "limiting values" is oscillation (F), the "word problem" is G, and the "exam twist" is H. Every cell gets hit below.

The figure below is your visual index to this matrix — a gallery of the four shapes these cells produce. Study it before reading the examples:

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Example 1 — Cell A: the $0/0$ that cancels (removable)

Forecast: Plugging $x=3$ gives top $=9-3-6=0$ and bottom $=0$. A $0/0$. Guess now: does the top share the factor $(x-3)$? If yes → removable hole. If no → infinite wall.

1. Compute $f(3)$. Top $=3^2-3-6=0$, bottom $=3-3=0$, so $f(3)=0/0$ — undefined. Condition V fails. Why this step? Before touching limits we always test whether the point even exists; here it does not.
2. Factor the top. $x^2-x-6=(x-3)(x+2)$. So for $x\ne 3$, $f(x)=\frac{(x-3)(x+2)}{x-3}=x+2.$ Why this step? The limit ignores the single point $x=3$; cancelling the common $(x-3)$ reveals the value the graph is aiming at. This is legal precisely because $x\ne 3$ near the point.
3. Take the limit. $\underset{x\to 3}{\lim }(x+2)=5$, finite, and left $=$ right (it's a polynomial). Condition L holds. Why this step? Removable-vs-infinite is decided by whether the limit is finite; a polynomial has one obvious two-sided value, so we read it off directly.
4. Compare. Limit $=5$ exists but $f(3)$ is undefined → removable. Patch: define $f(3)=5$. Why this step? Classification is precisely the comparison of L against V; here L exists (finite) but V is missing, which is the definition of removable.

Verify: approach from $x=2.99$: $x+2=4.99$; from $x=3.01$: $5.01$. Both squeeze onto $5$. Single missing dot → removable. ✓

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Example 2 — Cell B: the $0/0$ that does NOT cancel (infinite)

Forecast: Again top and bottom both vanish at $3$. But watch the bottom: $x^2-6x+9=(x-3{)}^2$ — a squared factor. After one cancel, a $(x-3)$ still hides in the denominator. Guess: wall, not hole.

1. Compute $f(3)$. Top $=0$, bottom $=(3-3{)}^2=0$, so $0/0$ — undefined. V fails. Why? Same first check as always.
2. Factor and cancel once. $f(x)=\frac{x-3}{(x-3{)}^2}=\frac{1}{x-3},x\ne 3.$ Why this step? We cancel exactly as many $(x-3)$'s as both sides share. Here the bottom had two and the top one, so one survives downstairs — the tell-tale sign of a surviving wall.
3. Take one-sided limits (see Vertical asymptotes and rational functions).
   • From the left, $x=2.999$: $x-3=-0.001$, so $1/(x-3)=-1000\to -\infty$.
   • From the right, $x=3.001$: $x-3=+0.001$, so $1/(x-3)=+1000\to +\infty$. Why split into sides? Because the sign of the tiny denominator flips as we cross $3$; a two-sided statement would hide that.
4. Compare. A side is infinite → infinite discontinuity. This is the parent-note mistake in action: undefined ≠ removable. Why this step? Classification always ends by matching what we found against the parent-note definitions; an infinite one-sided limit is exactly the "infinite" case (the parent note's "pole" or vertical-asymptote type).

Verify: $f(2.999)=-1000$, $f(3.001)=+1000$ — magnitudes explode, signs oppose → genuine pole (infinite discontinuity). ✓

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Example 3 — Cell C: piecewise that is actually continuous

Forecast: Two different formulas meet at $x=1$. Guess: do the two heights match there? If yes, the pen never lifts.

1. Value. $x=1$ uses $x\ge 1\Rightarrow x^2+2$, so $f(1)=1^2+2=3$. V holds. Why the second piece? The inequality is $x\ge 1$, and $1$ satisfies "$\ge 1$". (Parent mistake: pick by the exact inequality, not by proximity.)
2. Left limit. $L^{-}={\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3$. Why this piece? Coming from below, $x<1$, which uses $2x+1$.
3. Right limit. $L^{+}={\lim }_{x\to 1^{+}}(x^2+2)=1+2=3$. Why this piece? Coming from above, $x\ge 1$, which uses $x^2+2$.
4. Compare. $L^{-}=L^{+}=f(1)=3$ — all three equal → continuous. No discontinuity at all. Why this step? Continuity is exactly the VLE triple agreeing; we only declare "continuous" once we have literally checked all three numbers match.

Verify: $L^{-}=3$, $L^{+}=3$, $f(1)=3$ — the VLE triple all agree. ✓ See Piecewise functions for why matching heights at a seam is exactly the continuity condition.

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Example 4 — Cell D: piecewise that jumps

Forecast: Almost Example 3, but the second piece is now $x^2+5$. Guess: the right side lands higher — a step.

1. Value. $f(1)=1^2+5=6$. V holds. Why this piece? $x=1$ satisfies $x\ge 1$, so we use $x^2+5$.
2. Left limit. $L^{-}=2(1)+1=3$. Why this piece? Approaching from below uses $x<1\Rightarrow 2x+1$.
3. Right limit. $L^{+}=1^2+5=6$. Why this piece? Approaching from above uses $x\ge 1\Rightarrow x^2+5$.
4. Compare. $L^{-}=3\ne 6=L^{+}$, both finite → jump. Jump size $=|L^{+}-L^{-}|=|6-3|=3$. Why "jump" not "removable"? Removable needs $L^{-}=L^{+}$ (limit exists). Here the two sides disagree, so no single dot can fill the gap.

Verify: left approaches $3$, right approaches $6$; gap $=3$. Look at the figure: the pink left-piece open dot sits at height $3$, the blue right-piece filled dot at $6$. ✓

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Example 5 — Cell E: a wall where BOTH sides go the same way

Forecast: In Example 2 the two sides shot to $-\infty$ and $+\infty$. Here the denominator is squared, so it is always positive. Guess: both sides rocket the same way (up).

1. Value. $f(2)=1/0$ — undefined. V fails. Why this step? We always start with the value check; a $1/0$ tells us the point does not exist and warns of a possible wall.
2. Analyse the sign of the denominator. $(x-2{)}^2\ge 0$ for every $x$, and it is a tiny positive number near $2$ (never negative). Why does sign matter? A positive tiny denominator makes $1/\text{tiny}$ a large positive number regardless of which side we come from — the squared power removes the sign flip we saw in Ex 2.
3. One-sided limits.
   • Left, $x=1.99$: $(x-2{)}^2=0.0001$, so $f=10000\to +\infty$.
   • Right, $x=2.01$: $(x-2{)}^2=0.0001$, so $f=10000\to +\infty$. Why compute both sides? An infinite discontinuity only needs one side to blow up; checking both confirms it is a two-sided upward wall, not a one-sided oddity.
4. Compare. Both sides $=+\infty$ → infinite discontinuity (a two-sided upward asymptote / pole). Why this step? Matching our findings to the parent definitions: any side $=\pm \infty$ is the infinite case.

Verify: $f(1.99)=10000=f(2.01)$; both blow up positively — contrast this with Ex 2 where the odd power flipped the sign. ✓

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Example 6 — Cell F: oscillation that never settles

Forecast: As $x\to 0$, the inside $1/x$ races to $\pm \infty$, so $\sin$ swings between $-1$ and $1$ faster and faster. Guess: the graph never approaches one number — the limit simply does not exist, and not because of infinity.

1. Value. $f(0)=\sin (1/0)$ — undefined. V fails. Why this step? The usual first check: $1/0$ inside means the point does not exist.
2. Test whether the limit exists. Take two shrinking sequences that go to $0$:
   • $x_n=\frac{1}{2\pi n}$ gives $\sin (2\pi n)=0$.
   • $x_n=\frac{1}{2\pi n+\pi /2}$ gives $\sin (2\pi n+\pi /2)=1$. Why two sequences? If a limit existed, every way of approaching $0$ would give the same value. We found two approaches giving $0$ and $1$ — so no single limit exists.
3. Classify. The one-sided limits are neither finite-and-equal (not removable) nor $\pm \infty$ (not the simple pole) — the function oscillates. Why this step? We rule out each named type in turn: not removable (no finite limit), not jump (no two finite one-sided values), not infinite (it does not blow up) — what remains is the oscillating case.

Verify: on the sequence $x_n=1/(2\pi n)$ the value is $0$; on $x_n=1/(2\pi n+\pi /2)$ the value is $1$. Two limits ≠ one → no limit. ✓

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Example 7 — Cell G: real-world word problem (parking meter)

Forecast: Cost is constant, then jumps up by ₹20 at each whole hour. Guess: jump discontinuity at $t=1$, size ₹20.

1. Value at $t=1$. The clause $0<t\le 1$ includes $t=1$, so $C(1)=30$. V holds. Why this clause? $t=1$ satisfies "$\le 1$", so the ₹30 rate still applies exactly at one hour.
2. Left limit. For $t$ just below $1$, still in $0<t\le 1$: $L^{-}=30$. Why this piece? Values just under one hour are billed at the flat ₹30 rate.
3. Right limit. For $t$ just above $1$, we enter $1<t\le 2$: $L^{+}=50$. Why this piece? The instant we pass one hour, the second-hour band ₹50 applies.
4. Compare. $L^{-}=30\ne 50=L^{+}$, both finite → jump. Jump size $=|50-30|=20$, i.e. ₹20 — exactly the extra hour's charge. Why this step? Two finite but unequal one-sided values is precisely the jump definition; the gap size is what the customer is charged crossing the boundary.

Verify (with units): left-approach cost ₹30, right-approach cost ₹50, difference ₹20 = the stated second-hour fee. The model should jump — real fees step, they don't ramp. ✓

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Example 8 — Cell H: the exam twist (find $k$)

Forecast: The top piece has a removable hole (Ex-1 style). Guess: $k$ must equal whatever the pieces limit to — plug the hole with the right dot.

1. Value. By definition $f(3)=k$ (that's the whole point — $k$ is ours to choose). V holds for any $k$. Why this step? We must know the actual value at $3$ before we can force it to equal the limit; the definition hands it to us as $k$.
2. Find the limit of the other piece. For $x\ne 3$, $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3,$ so $\underset{x\to 3}{\lim }f(x)=3+3=6$. Why this step? Factor–cancel reveals the value the curve aims for at $3$ (the hole); this is the target $k$ must hit.
3. Force condition E. Continuity needs limit $=f(3)$, i.e. $6=k$. So $\boxed{k=6}$. Why solve this equation? We don't test here — we design. Setting $k$ equal to the limit closes the pen-lift.

Verify: with $k=6$: $L^{-}=L^{+}=6$ (the polynomial $x+3$) and $f(3)=6$ — all three agree, continuous. Any other $k$ (say $k=5$) would make it removable-again with a misplaced dot. ✓

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Recall Self-test before you move on

A rational function gives $0/0$ but the denominator had a squared factor. Type? ::: Infinite (a pole) — one $(x-a)$ survives downstairs (Cell B/E). $f$ jumps from $3$ to $6$ at $x=1$. Jump size? ::: $|6-3|=3$. Why is $\sin (1/x)$ at $0$ not removable? ::: The limit does not exist (it oscillates), and removable requires a finite limit — so it is essential/second-kind. To make $\frac{x^2-9}{x-3}$ continuous at $3$, set $k=$ ? ::: $6$. In Cell E, why do both sides go to $+\infty$? ::: The denominator is squared, so it is always positive near the point.

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Connections

• Continuity — definition, types of discontinuity (removable, jump, infinite) (parent — the VLE test and type definitions)
• Limits — definition and one-sided limits (every example uses $L^{-}$ and $L^{+}$)
• Vertical asymptotes and rational functions (Cells B, E — the surviving-factor walls / poles)
• Piecewise functions (Cells C, D, G, H — seams and unknown constants)