Worked examples — Intuitive concept of a limit — table of values, graphical

Before anything, recall the one rule that governs all of this: a limit watches where $f(x)$ is heading as $x$ creeps toward $a$ from both sides, and it ignores what happens at $a$ itself. Keep that in your pocket.

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The scenario matrix

Every limit-at-a-point problem falls into one of these cells. Each row is a behaviour class; the last column names the example that conquers it.

#	Case class	What you see when you look	Trap	Example
A	Direct substitution works (continuous)	Curve passes smoothly through the point	none	Ex 1
B	$0/0$ hole, factor & cancel	Line with a missing dot	can't plug in	Ex 2
C	$0/0$ needing a rationalising trick (roots)	Smooth curve, hidden hole	factoring won't help	Ex 3
D	Two-sided limit fails — a jump	Two pieces at different heights	one side isn't enough	Ex 4
E	Blow-up to infinity ($k/0$, $k\ne 0$)	Curve rockets up/down	"$=\infty$" is not a number	Ex 5
F	Sign matters — left and right go opposite infinities	One branch dives, one soars	signs of the denominator	Ex 5
G	Limit value $\ne$ function value (stray dot)	Curve heading one place, a dot placed elsewhere	trusting $f(a)$	Ex 6
H	Wild oscillation — no destination	Curve wobbles forever near $a$	assuming a limit must exist	Ex 7
I	Real-world word problem	An average speed heading toward instantaneous	interpreting "approaches"	Ex 8
J	Exam twist — limit at infinity	Curve flattening toward a shelf	large-$x$ dominance	Ex 9

We now clear the board, one cell at a time.

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Ex 1 — Cell A: substitution just works

Forecast: this is a plain polynomial with no fractions and no roots — nothing can break. Guess the value before reading on: what do you get if you simply put $x=2$?

Steps.

1. Put $x=2$ into the expression: $3(2{)}^2-2+4$. Why this step? For a polynomial there are no divisions by zero and no undefined operations anywhere, so the curve is a single unbroken line of ink — its destination equals its actual value. (This is the idea of Continuity at a point: for such functions, limit = plug-in.)
2. Compute: $3\cdot 4-2+4=12-2+4=14$.

${\lim }_{x\to 2}(3x^2-x+4)=14$

Verify: test a value just short of $2$, say $x=1.99$: $3(1.99{)}^2-1.99+4=3(3.9601)-1.99+4=11.8803+2.01=13.8903$, hugging $14$. ✓

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Ex 2 — Cell B: the $0/0$ hole, cured by factoring

Forecast: put $x=3$ and you get $\frac{9-9}{3-3}=\frac{0}{0}$ — a trap, not an answer. But the top clearly wants to factor. Where do you think the curve is heading?

Steps.

1. Notice $x^2-9=(x-3)(x+3)$ (difference of two squares). Why this step? The reason we got $0/0$ is that both top and bottom secretly contain the factor $(x-3)$; factoring exposes it so we can deal with the common cause.
2. Cancel the shared $(x-3)$: $\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3).$ Why this step? Cancelling is legal because $x\to 3$ means $x$ is near but never equal to $3$, so $x-3\ne 0$ and dividing by it is allowed. Away from the single point $x=3$, the function is just the tame line $x+3$.
3. Now the substitution trap is gone — take the limit of $x+3$ at $3$: $3+3=6$.

${\lim }_{x\to 3}\frac{x^2-9}{x-3}=6$

Verify: table check at $x=2.999$: $\frac{2.999^2-9}{2.999-3}=\frac{8.994001-9}{-0.001}=\frac{-0.005999}{-0.001}=5.999$, aiming at $6$. ✓

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Ex 3 — Cell C: $0/0$ with a root, cured by rationalising

Forecast: plug in $x=0$: $\frac{\sqrt{4}-2}{0}=\frac{0}{0}$. Factoring won't help — the culprit is a square root. Guess: does the answer feel small (like a fraction) or large?

Steps.

1. Multiply top and bottom by the conjugate $\sqrt{x+4}+2$: $\frac{\sqrt{x+4}-2}{x}\cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}.$ Why this step? Why the conjugate and not factoring? Because $(\sqrt{u}-2)(\sqrt{u}+2)=u-4$ erases the root on top. We use this specific tool because a difference of square roots is exactly what "difference of squares" was built to flatten.
2. The numerator becomes $(x+4)-4=x$: $\frac{x}{x\left(\sqrt{x+4}+2\right)}.$ Why this step? This is our reward — the offending $x$ that caused the $0/0$ now appears as a clean common factor.
3. Cancel the $x$ (legal since $x\to 0$ means $x\ne 0$): $\frac{1}{\sqrt{x+4}+2}.$
4. Substitute $x=0$: $\frac{1}{\sqrt{4}+2}=\frac{1}{2+2}=\frac{1}{4}$.

${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}=\frac{1}{4}$

Verify: at $x=0.001$: $\frac{\sqrt{4.001}-2}{0.001}=\frac{2.00024998-2}{0.001}=\frac{0.00024998}{0.001}=0.24998\approx 0.25$. ✓ (This trick is the heart of Indeterminate forms and algebraic simplification.)

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Ex 4 — Cell D: a jump, so the two-sided limit fails

Forecast: $|x|$ is "$x$ made positive." Dividing by $x$… what happens for negatives versus positives? Guess whether one clean number appears.

Steps.

1. For $x>0$, $|x|=x$, so $f(x)=\frac{x}{x}=1$. Right-hand limit $\underset{x\to 0^{+}}{\lim }f(x)=1$. Why this step? On the right branch (the red horizontal segment in the figure) the function is a flat line at height $1$.
2. For $x<0$, $|x|=-x$, so $f(x)=\frac{-x}{x}=-1$. Left-hand limit $\underset{x\to 0^{-}}{\lim }f(x)=-1$. Why this step? On the left branch (black segment) the function sits flat at $-1$. Look at the figure: the two segments approach $x=0$ from different heights.
3. Compare: left gives $-1$, right gives $+1$. They disagree.

${\lim }_{x\to 0}\frac{|x|}{x}\text{does not exist.}$

Verify: $x=-0.001\Rightarrow \frac{0.001}{-0.001}=-1$; $x=+0.001\Rightarrow \frac{0.001}{0.001}=1$. Two different targets ⇒ no single limit. ✓ (More in One-sided limits and limits at infinity.)

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Ex 5 — Cells E & F: blow-up, and the two sides going opposite infinities

Forecast: at $x=2$ both denominators are $0$ but the tops are $1$, not $0$ — so this is not $0/0$. It's "a nonzero number over something shrinking to zero," which explodes. But does it explode up or down? That depends on the sign of the denominator.

Part I — $\frac{1}{x-2}$ (Cell F, opposite signs).

1. Approach from the right ($x\to 2^{+}$): then $x-2$ is a tiny positive number, so $\frac{1}{x-2}$ is a huge positive number. $\underset{x\to 2^{+}}{\lim }\frac{1}{x-2}=+\infty$. Why this step? Dividing $1$ by something like $0.001$ gives $1000$; by $0.0001$ gives $10000$ — no ceiling. The red branch in the figure shoots up.
2. Approach from the left ($x\to 2^{-}$): then $x-2$ is a tiny negative number, so $\frac{1}{x-2}$ is a huge negative number. $\underset{x\to 2^{-}}{\lim }\frac{1}{x-2}=-\infty$. Why this step? The black branch dives down. Because the two sides run to opposite infinities, the two-sided limit does not exist (and isn't even "$\infty$").

Part II — $\frac{1}{(x-2{)}^2}$ (Cell E, same infinity).

3. Now the denominator $(x-2{)}^2$ is a square, so it is positive from both sides. Both one-sided limits go to $+\infty$. Why this step? Squaring removes the sign issue; from either side we divide $1$ by a tiny positive number. Here we may write $\underset{x\to 2}{\lim }\frac{1}{(x-2{)}^2}=+\infty$ because both sides agree on the same runaway direction.

Verify: $\frac{1}{2.001-2}=1000$ (up) but $\frac{1}{1.999-2}=-1000$ (down) → sides disagree. For the square: $\frac{1}{(2.001-2{)}^2}=10^6$ and $\frac{1}{(1.999-2{)}^2}=10^6$ → both huge positive. ✓

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Ex 6 — Cell G: the stray dot (limit ≠ value)

Forecast: the definition parks a random dot at height $7$ when $x=1$. But the limit only watches the approach — does that dot matter?

Steps.

1. For every $x\ne 1$ the rule is $x^2+1$, and the limit never lets $x$ equal $1$. Why this step? Since the limit ignores the single point $x=1$, we simply use the formula that governs all the nearby points.
2. Take the limit of $x^2+1$ at $1$: $1^2+1=2$.
3. But the defined value is $g(1)=7$.

${\lim }_{x\to 1}g(x)=2,g(1)=7,2\ne 7.$

The curve heads for height $2$; the lone dot at $7$ is irrelevant to the limit.

Verify: at $x=0.999$: $0.999^2+1=1.998001$, and at $x=1.001$: $1.001^2+1=2.002001$ — both squeezing toward $2$, not $7$. ✓

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Ex 7 — Cell H: endless oscillation, no destination

Forecast: as $x\to 0$, the fraction $\frac{1}{x}$ becomes enormous, so the angle inside the sine spins faster and faster. Does the output ever settle?

Steps.

1. Understand $\frac{1}{x}$ near $0$: for $x=0.1$ it's $10$; for $x=0.01$ it's $100$; for $x=0.001$ it's $1000$. It races to infinity. Why this step? The behaviour of $\sin$ depends entirely on its input angle, so we must first see how fast that angle grows.
2. $\sin$ of an ever-growing angle keeps sweeping the full range $[-1,1]$, over and over, infinitely many times in any tiny window around $0$. Why this step? Look at the red curve in the figure: near $x=0$ it is a solid band of oscillation — there is no single height it approaches.
3. Because we can find points arbitrarily close to $0$ where the value is $+1$ and other points equally close where it is $-1$, no single number $L$ is being approached.

${\lim }_{x\to 0}\sin \left(\frac{1}{x}\right)\text{does not exist.}$

Verify: at $x=\frac{2}{\pi }$-type inputs the value is $1$, at others $-1$. Take $x_1=\frac{1}{5000\pi }$ (value $0$) and $x_2=\frac{2}{(4k+1)\pi }$ (value $1$), both near $0$ — different heights ⇒ no limit. ✓ (Compare with the Squeeze (Sandwich) Theorem: multiplying this by $x$ would force a limit, because it clamps the wobble.)

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Ex 8 — Cell I: real-world word problem (average → instantaneous)

Forecast: as the time window $h$ shrinks to zero, the average speed should home in on the "speedometer reading" at the instant $t=2$. Guess: bigger or smaller than $20$?

Steps.

1. Compute the pieces: $s(2+h)=5(2+h{)}^2=5(4+4h+h^2)=20+20h+5h^2$, and $s(2)=5(4)=20$. Why this step? We need the change in distance over the window before we can average it.
2. Form the average speed: $v_{\text{avg}}(h)=\frac{(20+20h+5h^2)-20}{h}=\frac{20h+5h^2}{h}.$ Why this step? Direct substitution $h=0$ gives $\frac{0}{0}$ — the same trap as before, because the window has zero width. So we simplify instead.
3. Cancel $h$ (legal since $h\to 0$ means $h\ne 0$): $v_{\text{avg}}(h)=20+5h$.
4. Take the limit: $\underset{h\to 0}{\lim }(20+5h)=20$.

$\text{Instantaneous speed at }t=2\text{ s}=20\text{m/s}.$

Verify (units + number): $5t^2$ has metres; dividing metres by seconds gives m/s ✓. At $h=0.001$: $20+5(0.001)=20.005$ m/s, hugging $20$. ✓ (This is the seed of The derivative as a limit of difference quotients.)

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Ex 9 — Cell J: exam twist — limit at infinity

Forecast: now $x$ runs to infinity instead of to a point. Both top and bottom explode, so it's an $\infty /\infty$ tug-of-war. The winner is decided by the highest powers. Guess the shelf height.

Steps.

1. Divide every term, top and bottom, by the highest power present, $x^2$: $\frac{3+\frac{5}{x}}{2-\frac{7}{x^2}}.$ Why this step? We use this dividing trick because as $x\to \infty$, any term of the form $\frac{\text{constant}}{x^k}$ shrinks to $0$ — so this rewrite isolates what survives from what vanishes.
2. As $x\to \infty$: $\frac{5}{x}\to 0$ and $\frac{7}{x^2}\to 0$. Why this step? Dividing a fixed number by an ever-larger $x$ drives it toward zero — the same "$k/\text{big}$" idea from Ex 5, now used to kill terms.
3. What remains: $\frac{3+0}{2-0}=\frac{3}{2}$.

${\lim }_{x\to \infty }\frac{3x^2+5x}{2x^2-7}=\frac{3}{2}$

Verify: at $x=1000$: $\frac{3\cdot 10^6+5000}{2\cdot 10^6-7}=\frac{3005000}{1999993}=1.50250\dots \approx 1.5$. ✓ (Full treatment in One-sided limits and limits at infinity.)

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Recall

Recall Which two cells give "does not exist," and why are they different?

Cell D (jump): the two sides disagree on a finite value. Cell H (oscillation): no value at all is approached. Cell F ($\frac{1}{x-2}$) also fails two-sided because the sides run to opposite infinities.

Recall When you hit

$0/0$, what are your two main tools and how do you choose? Factor & cancel (Ex 2) when it's a polynomial; multiply by the conjugate (Ex 3) when a square root is involved.

Recall Why is

${\lim }_{x\to 2}\frac{1}{(x-2{)}^2}=+\infty$ allowed as an answer but ${\lim }_{x\to 2}\frac{1}{x-2}$ is not? The square keeps the denominator positive on both sides, so both agree on $+\infty$; the plain version gives $+\infty$ on the right but $-\infty$ on the left — the sides disagree.

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Connections

• Intuitive concept of a limit — table of values, graphical
• One-sided limits and limits at infinity
• Indeterminate forms and algebraic simplification
• Continuity at a point
• The derivative as a limit of difference quotients
• Squeeze (Sandwich) Theorem
• Formal epsilon-delta definition of a limit