Functions, limits, and continuity: Worked examples

The scenario matrix

Every limit-and-continuity problem you will meet is one of these cells. We hit each one at least once.

Cell	What makes it special	What you DO	Example
A. Nice / plug-in	function continuous at the point	just substitute	Ex 1
B. Removable $0/0$	numerator & denominator both $\to 0$ , factor cancels	factor & cancel, then plug in	Ex 2
C. Rationalize $0/0$	$0/0$ with a square root, can't factor directly	multiply by conjugate	Ex 3
D. Jump (one-sided)	left limit $\neq =$ right limit	compute both sides separately	Ex 4
E. Infinite / vertical asymptote	denominator $\to 0$ but numerator doesn't	check sign from each side, $\pm \infty$	Ex 5
F. Limit at infinity	$x \to + \infty$ and $x \to - \infty$ (end behaviour)	divide by highest power, watch signs	Ex 6
G. Continuity by design	choose a constant to remove a hole	set piece value $=$ limit	Ex 7
H. ML kink (ReLU)	continuous but not differentiable	check HEA vs slope	Ex 8
I. Word problem	real quantity, must interpret + units	model, then take limit	Ex 9
J. Exam twist	looks like $0/0$ but is really a jump	test both sides — trap!	Ex 10
K. Other indeterminates	$0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$	rewrite into $0/0$ or $\infty/\infty$ first	Ex 11

We use one repeated tool: factoring turns a $0/0$ into something you can plug in — because the limit never touches the forbidden point, only its neighbourhood. Keep that in mind throughout.

Ex 1 — Cell A: the polite plug-in

Ex 2 — Cell B: the removable $0/0$ hole

Worked example Compute

$x \to 3 lim \frac{x ^{2} - 9}{x - 3}$ . Forecast: plug in $x = 3$ and you get $\frac{0}{0}$ . That is not the answer — guess the real one.

Recognise the form: $9 - 9 = 0$ on top, $3 - 3 = 0$ on bottom → indeterminate $0/0$ . Why this step? $0/0$ is a signal to simplify, never a verdict (parent §4, Mistake #2).
Factor the top: $x^{2} - 9 = (x - 3) (x + 3)$ . So $\frac{( x - 3 ) ( x + 3 )}{x - 3}$ . Why this step? We hunt for the same factor top and bottom because that shared factor is what forces the $0/0$ .
Cancel $x - 3$ : for $x \neq = 3$ the expression equals $x + 3$ . Why this step? The limit only looks near $x = 3$ , never at it, so $x - 3 \neq = 0$ and cancelling is legal.
Now plug in the survivor: $3 + 3 = 6$ . Why this step? After cancelling, $x + 3$ is a polynomial — continuous everywhere — so by condition 3 (A) substitution equals the limit.

Picture: the graph is the straight line $y = x + 3$ with a single hole punched at $(3, 6)$ — see figure.

Verify: at $x = 2.999$ , $\frac{x ^{2} - 9}{x - 3} = 5.999$ ; at $x = 3.001$ , $= 6.001$ . Heading to $6$ . ✅

Ex 3 — Cell C: rationalize the $0/0$

Worked example Compute

$x \to 0 lim \frac{x + 4 - 2}{x}$ . Forecast: $x = 0$ gives $\frac{4 - 2}{0} = \frac{0}{0}$ . There's no obvious factor. Guess anyway.

Spot $0/0$ with a square root — factoring won't cut it. Why this step? Roots hide their "zero" inside; the tool that exposes it is the conjugate $x + 4 + 2$ .
Multiply top and bottom by that conjugate: $\frac{x + 4 - 2}{x} \cdot \frac{x + 4 + 2}{x + 4 + 2} = \frac{( x + 4 ) - 4}{x ( x + 4 + 2 )} .$ Why this step? $(a - b) (a + b) = a^{2} - b^{2}$ kills the root in the numerator, turning it into plain algebra.
Simplify the top: $(x + 4) - 4 = x$ . So we have $\frac{x}{x ( x + 4 + 2 )}$ . Why this step? Collapsing the top to a bare $x$ exposes the same factor $x$ sitting in the bottom — that shared $x$ is what was forcing the $0/0$ .
Cancel $x$ : for $x \neq = 0$ the expression equals $\frac{1}{x + 4 + 2}$ . Why this step? The limit only looks near $x = 0$ , never at it, so $x \neq = 0$ and cancelling is legal — exactly the same trick as Ex 2.
Plug in $x = 0$ : $\frac{1}{4 + 2} = \frac{1}{4}$ . Why this step? The survivor $\frac{1}{x + 4 + 2}$ is continuous at $0$ (denominator $= 4 \neq = 0$ ), so substitution equals the limit.

Verify: at $x = 0.001$ , $\frac{4.001 - 2}{0.001} \approx 0.24998 \approx \frac{1}{4}$ . ✅

Ex 4 — Cell D: a genuine jump (one-sided limits)

Worked example For the step function

$f (x) = {x + 1 x - 1 x < 0 x \geq 0$ , find $lim_{x \to 0^{-}} f$ , $lim_{x \to 0^{+}} f$ , and $lim_{x \to 0} f$ . Forecast: do the two sides agree?

Left limit $lim_{x \to 0^{-}}$ : use the $x < 0$ rule, $f (x) = x + 1 \to 0 + 1 = 1$ . Why this step? Approaching from the left ( $x < 0$ ) we live entirely in the top branch.
Right limit $lim_{x \to 0^{+}}$ : use the $x \geq 0$ rule, $f (x) = x - 1 \to 0 - 1 = - 1$ . Why this step? Approaching from the right ( $x > 0$ ) we live entirely in the bottom branch — a piecewise function must be evaluated with the branch that owns the side you're coming from.
Compare: left heading $= 1$ , right heading $= - 1$ . They disagree. Why this step? A two-sided limit exists only if both sides funnel to the same number (parent §2).
Conclusion: $lim_{x \to 0} f (x)$ does not exist — this is a jump discontinuity of size $2$ . Why this step? Because condition 2 (E) fails (no shared heading), the two-sided limit can't exist, and no amount of redefining $f (0)$ can fix a jump — the gap is the size of the leap.

Verify: $f (- 0.001) = 0.999 \to 1$ , $f (0.001) = - 0.999 \to - 1$ . Gap $= 2$ , no shared heading. ✅

Ex 5 — Cell E: the explosion (infinite limit, both signs)

Worked example Compute

$x \to 2^{-} lim \frac{1}{x - 2}$ and $x \to 2^{+} lim \frac{1}{x - 2}$ . Forecast: the denominator $\to 0$ but the top is fixed at $1$ . Does it blow up — and which way?

Note the numerator is $1$ (not zero), denominator $\to 0$ . This is not $0/0$ ; it's "finite over zero" → an infinite limit. Why this step? Only when denominator $\to 0$ AND numerator $\to 0$ is it indeterminate. Here it's a vertical asymptote.
From the left $x \to 2^{-}$ : take $x = 1.9$ , then $x - 2 = - 0.1 < 0$ , so $\frac{1}{- 0.1} = - 10$ . Closer, $x = 1.99 \Rightarrow - 100$ . It dives to $- \infty$ . Why this step? Sign of the denominator decides the sign of the explosion — you must check each side.
From the right $x \to 2^{+}$ : $x = 2.1 \Rightarrow x - 2 = 0.1 > 0$ , $\frac{1}{0.1} = + 10$ ; $x = 2.01 \Rightarrow + 100$ . It rockets to $+ \infty$ . Why this step? On the right the denominator is positive and shrinking, so $1/ (tiny positive)$ grows without bound upward — the mirror image of the left side.
Conclusion: $lim_{x \to 2^{-}} = - \infty$ , $lim_{x \to 2^{+}} = + \infty$ . The two-sided limit does not exist (infinite, opposite signs). Why this step? Both condition 1 (H) ( $f (2)$ undefined) and condition 2 (E) fail; naming the sign on each side is what tells you it's an infinite discontinuity, not a jump.

Verify: $\frac{1}{1.99 - 2} = - 100$ , $\frac{1}{2.01 - 2} = 100$ . ✅

Ex 6 — Cell F: end behaviour, BOTH directions

Worked example Compute

$x \to + \infty lim \frac{3 x ^{2} + 5 x}{2 x ^{2} - 7}$ and $x \to - \infty lim \frac{3 x ^{2} + 5 x}{2 x ^{2} - 7}$ . Forecast: both top and bottom blow up. Ratio of two infinities — what wins, and does the sign of $x$ change the answer?

Both $\to \infty$ (in magnitude), so it's $\infty/\infty$ — another indeterminate form. Why this step? Like $0/0$ , this needs simplifying, not guessing.
Divide every term by the highest power present, $x^{2}$ : $\frac{3 + \frac{5}{x}}{2 - \frac{7}{x ^{2}}} .$ Why this step? Dividing by the dominant power reveals which terms fade. As $∣ x ∣ \to \infty$ , $\frac{5}{x} \to 0$ and $\frac{7}{x ^{2}} \to 0$ .
As $x \to + \infty$ : $\frac{3 + 0}{2 - 0} = \frac{3}{2}$ . Why this step? Only the leading coefficients survive — the horizontal asymptote is the ratio $3/2$ .
As $x \to - \infty$ : the same terms $\frac{5}{x}$ and $\frac{7}{x ^{2}}$ still $\to 0$ (a negative-then-tiny number is still tiny), so again $\frac{3 + 0}{2 - 0} = \frac{3}{2}$ . Why this step? Here the leading powers both are even ( $x^{2}$ ), so sign of $x$ doesn't flip anything — the curve approaches the same horizontal line $y = \frac{3}{2}$ on both ends. (If the top were degree-1 like $3 x$ , the two ends would differ in sign — always check.)

Verify: at $x = 1000$ and $x = - 1000$ , $\frac{3 x ^{2} + 5 x}{2 x ^{2} - 7} \approx 1.5006$ and $\approx 1.4994$ — both heading to $\frac{3}{2}$ . ✅

Ex 7 — Cell G: continuity by design (choose the constant)

Ex 8 — Cell H: the ML kink (continuous, not differentiable)

Worked example Is

$ReLU (x) = max (0, x)$ continuous at $x = 0$ ? Is it differentiable there? Forecast: parent §3 hinted at this — recall the answer before checking.

Condition 1 (H): $ReLU (0) = max (0, 0) = 0$ . Exists. ✅ Why this step? HEA condition 1 asks only whether the machine produces an output at $x = 0$ — it does, namely $0$ .
Condition 2 (E): from the left $x < 0$ , $ReLU = 0 \to 0$ ; from the right $x > 0$ , $ReLU = x \to 0$ . Both sides $\to 0$ → limit exists, equals $0$ . Why this step? Unlike the jump in Ex 4, here both branches meet at $0$ , so no jump.
Condition 3 (A): $lim_{x \to 0} = 0 = ReLU (0)$ . Equal → continuous. ✅ Why this step? Heading ( $0$ ) equals arrival ( $0$ ), so the third HEA condition passes and ReLU is continuous at $0$ .
Differentiability: the slope from the left is $0$ (flat), from the right is $1$ (line $y = x$ ). $0 \neq = 1$ : the slopes disagree — a corner, no unique tangent. Why this step? A derivative is itself a limit (of slopes); if left-slope $\neq =$ right-slope, that limit fails. So not differentiable at $0$ .

This is why ReLU trains fine (continuous → loss doesn't jump) yet needs a subgradient at $0$ (see Gradient Descent). Continuity is weaker than differentiability.

Verify: left slope $\frac{ReLU ( - 0.001 ) - 0}{- 0.001} = 0$ ; right slope $\frac{ReLU ( 0.001 ) - 0}{0.001} = 1$ . Unequal. ✅

Ex 9 — Cell I: word problem with units

Worked example A drone's battery drains so that after

$t$ minutes the remaining charge is $C (t) = \frac{100 t}{t + 4}$ percent. What charge does the model predict it heads toward as flight time grows without bound? Forecast: does it approach $100%$ , some ceiling, or fall forever?

We want $lim_{t \to \infty} C (t)$ — the long-run heading, not any one moment. Why this step? "As time grows without bound" is precisely a limit at infinity (Cell F machinery).
Divide top and bottom by $t$ : $\frac{100}{1 + \frac{4}{t}}$ . Why this step? $\frac{4}{t} \to 0$ , exposing the dominant behaviour.
Limit: $\frac{100}{1 + 0} = 100$ . Why this step? With the fading term gone, only the constant ratio survives — this is the horizontal asymptote the reading approaches.
Interpret with units: $C$ is in percent, so the reading heads toward $100%$ — a horizontal asymptote. Why this step? A limit is just a bare number until you re-attach the physical unit; naming it " $100%$ " is what turns the algebra back into a statement about the drone. It never exceeds $100%$ (at $t = 4$ , $C = 50%$ ), which sanity-checks that this is a saturating quantity.

Verify: $C (4) = \frac{400}{8} = 50$ ; $C (1000) = \frac{100000}{1004} \approx 99.6%$ , climbing toward $100$ . ✅

Ex 10 — Cell J: the exam trap ( $∣ x ∣/ x$ )

Worked example Compute

$x \to 0 lim \frac{∣ x ∣}{x}$ . (Looks like $0/0$ — be careful.) Forecast: many students factor and answer " $1$ ". Test that instinct.

At $x = 0$ : $∣0∣/0 = 0/0$ . Tempting to "simplify" — but $∣ x ∣$ is a piecewise thing, not a polynomial you can factor. Why this step? The trap is treating $∣ x ∣$ like $x$ . It isn't: $∣ x ∣ = x$ for $x > 0$ but $∣ x ∣ = - x$ for $x < 0$ .
Right side $x \to 0^{+}$ ( $x > 0$ ): $\frac{∣ x ∣}{x} = \frac{x}{x} = 1$ . Why this step? On the right $∣ x ∣ = x$ (the positive branch), so the ratio collapses to $1$ — evaluate each branch with its own definition of $∣ x ∣$ .
Left side $x \to 0^{-}$ ( $x < 0$ ): $\frac{∣ x ∣}{x} = \frac{- x}{x} = - 1$ . Why this step? On the left $∣ x ∣ = - x$ (the negative branch), so numerator and denominator have opposite signs → ratio $- 1$ .
Left $= - 1 \neq = 1 =$ right. So the limit does not exist — it's a jump, disguised as a $0/0$ . Why this step? Condition 2 (E) fails: the two one-sided headings differ, so despite the $0/0$ appearance this is really a jump — the exam's whole trick.

Verify: $\frac{∣0.001∣}{0.001} = 1$ , $\frac{∣ - 0.001∣}{- 0.001} = - 1$ . No shared heading. ✅

Ex 11 — Cell K: the other indeterminate forms ( $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ )

Worked example Three quick classics that

look impossible until you rewrite them. Forecast: guess each before reading.

(a) $0 \cdot \infty$ : $x \to 0^{+} lim x ln x$ .

As $x \to 0^{+}$ , $x \to 0$ but $ln x \to - \infty$ — the form $0 \cdot (- \infty)$ , indeterminate. Why this step? A product where one factor dies and the other explodes has no automatic winner — you must convert it.
Rewrite as a quotient: $x ln x = \frac{ln x}{1/ x}$ , now the form $\frac{- \infty}{+ \infty}$ . Why this step? Turning $0 \cdot \infty$ into $\infty/\infty$ lets us use the ratio tools (or L'Hôpital's Rule) — you can only tackle a form you recognise.
By L'Hôpital's Rule (differentiate top and bottom): $\frac{1/ x}{- 1/ x ^{2}} = - x \to 0$ . Why this step? L'Hôpital is the licensed tool for $\infty/\infty$ ; the derivatives simplify to $- x$ , which plainly $\to 0$ .
So $lim_{x \to 0^{+}} x ln x = 0$ — the $x$ shrinking to zero beats the slow $ln$ blow-up.

(b) $\infty - \infty$ : $x \to \infty lim (x^{2} + x - x)$ .

Both pieces $\to \infty$ ; their difference is the indeterminate $\infty - \infty$ . Why this step? Two infinities racing to cancel — the gap between them can settle anywhere, so we must expose it.
Rationalize (same conjugate trick as Ex 3): multiply by $\frac{x ^{2} + x + x}{x ^{2} + x + x}$ to get $\frac{x}{x ^{2} + x + x}$ . Why this step? The conjugate turns the fragile difference into a clean quotient we can divide down.
Divide top and bottom by $x$ (Cell F move): $\frac{1}{1 + 1/ x + 1} \to \frac{1}{1 + 1} = \frac{1}{2}$ . Why this step? Dividing by the dominant power fades $1/ x \to 0$ , leaving the finite ratio $\frac{1}{2}$ .

(c) $1^{\infty}$ : $n \to \infty lim (1 + \frac{1}{n})^{n}$ .

Base $\to 1$ , exponent $\to \infty$ — the form $1^{\infty}$ , indeterminate (a base just above $1$ raised very high can land anywhere). Why this step? $1$ to any power looks like $1$ , but the base isn't exactly $1$ ; the tension between "almost 1" and "huge power" is real.
This is the definition of $e$ : $lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e \approx 2.71828$ . Why this step? Recognising a named limit saves the work — the compounding-interest limit is Euler's number, the backbone of the sigmoid and exponentials in ML.

Verify: (a) $0.01 ln 0.01 \approx - 0.046 \to 0$ ; (b) $1 0^{6} + 1000 - 1000 \approx 0.4999 \to \frac{1}{2}$ ; (c) $(1 + 1/10000)^{10000} \approx 2.71815 \to e$ . ✅

Recall Which cell am I in? (decision walkthrough)

Read this top to bottom — it's the same logic the diagram below draws, in words. Step 1 — plug in the point ::: If you get a clean number and the function is continuous → Cell A, done. Step 2 — got $0/0$ ? ::: Simplify — factor (B), rationalize if there's a root (C / Ex 11b), or use L'Hôpital's Rule. Step 3 — got finite $/0$ ? ::: Infinite limit — check the sign from each side → Cell E. Step 4 — piecewise, $∣ x ∣$ , or a step? ::: Compute both one-sided limits; if unequal → jump, DNE (D / J). Step 5 — $x \to \pm \infty$ ? ::: Divide by the highest power, check both ends → Cell F. Step 6 — $0 \cdot \infty$ , $\infty - \infty$ , or $1^{\infty}$ ? ::: Rewrite into $0/0$ or $\infty/\infty$ first, then apply Step 2's tools → Cell K. Step 7 — asked to choose a constant? ::: Set the piece value $=$ the limit → Cell G.

Connections

Derivatives and gradients — Ex 8's slope-mismatch is a failed derivative limit.
Gradient Descent — why the ReLU kink needs a subgradient.
Activation functions (ReLU, sigmoid) — continuity vs differentiability at the kink; the $e$ from Ex 11c powers the sigmoid.
L'Hôpital's Rule — the general tool for stubborn $0/0$ , $\infty/\infty$ , and the rewritten forms of Ex 11.
Continuity of composite functions — chaining continuous pieces (backprop backbone).

Scenario Map

The diagram below is just the [!recall] walkthrough drawn as arrows — plug in first; each way substitution can "break" sends you down a different branch to the matching cell.