Limits and Continuity

The derivative was defined as a limit of difference quotients. The integral was defined as a limit of Riemann sums. Neither unit said what a limit is.

That was serviceable while the limits behaved, $6+h$ visibly approaches 6, and sums bracketing $\frac{1}{3}$ visibly close on it. It stops being serviceable at the boundary cases those units raised and could not settle. Why do disagreeing one-sided quotients mean $|x|$ has no derivative at 0, rather than two? Why does $0/0$ carry no information when the difference quotient produces it every time? Why does an integrand blowing up inside the interval invalidate the fundamental theorem rather than merely complicating it?

Each answer is a fact about limits. This unit supplies the definition precise enough to settle them: a challenge-and-response formulation in which "approaches $L$" means every tolerance can be met.

With it come the limit laws, which make most limits a substitution; the indeterminate forms, which mark where substitution fails and algebra is required; the squeeze theorem, for limits no algebra reaches; and continuity, the condition that lets a limit be read off a function's value, together with the three ways it can fail, which are exactly the three ways the derivative failed.

The canonical intuition frames $\epsilon$–$\delta$ as a challenge and a response. Working one instance completely shows why the quantifiers are ordered as they are.

The claim. $\underset{x\to 3}{lim}(2x+1)=7$.

The challenge. Someone names a tolerance, say $\epsilon =0.01$: they want .

The response. Work backwards from what is wanted:

So the requirement is met exactly when . Answer $\delta =0.005$.

Verifying. At $x=2.995005$ and $x=3.004995$, just inside that $\delta$, the deviation $|f(x)-7|$ is $0.00999$, comfortably under $0.01$. The same rule answers every challenge:

Because a response exists for every $\epsilon$, the limit is 7.

Why the order of quantifiers matters. $\epsilon$ comes first and $\delta$ is chosen in response to it, $\delta$ is allowed to depend on $\epsilon$, and here it does. Reversing them would demand a single distance working for every tolerance, which nothing but a constant function could satisfy.

Why $x=a$ is excluded. The condition is , and the left inequality removes the point itself. This is not fastidiousness: it is what allows $\underset{x\to 2}{lim}\frac{x^2-4}{x-2}=4$ even though the function is undefined at 2. The limit asks where the function is heading, and a single missing or misplaced value cannot change that.

Why this is rarely done in practice. The definition justifies the shortcuts rather than replacing them. Once the limit laws are proved from it, each proof an $\epsilon$–$\delta$ argument, every limit built by arithmetic from continuous pieces is a substitution. The definition is called on only where substitution fails, which is precisely where the interesting cases live.

The canonical definition states the $\epsilon$–$\delta$ condition, the laws, continuity and the failure kinds. What follows is why each is phrased as it is.

Why continuity needs three conditions and not one. Writing "$f$ is continuous at $a$ when $\underset{x\to a}{lim}f(x)=f(a)$" looks sufficient but hides two prior requirements, and each fails independently:

Stating all three makes the diagnosis routine: check them in order and the first that fails names the defect.

Why the quotient law excludes a vanishing denominator. The law is proved by bounding $|\frac{f}{g}-\frac{L}{M}|$, and the bound divides by $|M|$. With $M=0$ there is nothing to divide by and the proof has no content, which is not a technicality, since $\frac{x^2-4}{x-2}$ at $x=2$ has a perfectly good limit of 4 that the law cannot deliver.

Why an indeterminate form is not a value. Three quotients all of the form $0/0$ as $x\to 0$: $\frac{x}{x}\to 1$, $\frac{x^2}{x}\to 0$, and $\frac{x}{x^2}\to \mathrm{\infty }$. Since the form is common to all three and the answers differ, the form carries no information. It is a signal that the laws do not apply, never a result.

Why removable and jump discontinuities are worth distinguishing. A removable discontinuity is a single misplaced or missing value: the limit exists, so redefining $f$ at that one point makes it continuous. A jump has no such repair. The one-sided limits genuinely differ, and no value assigned at the point can equal both. The distinction decides whether a theorem requiring continuity can be rescued by patching the function.

Why the squeeze theorem needs a common limit. If $g\le f\le h$ with $g\to L$ and $h\to M$ where $L\ne M$, then $f$ is confined to a band that never narrows, and nothing follows. The theorem works only when the band collapses to a point.

Why one-sided limits are enough for the two-sided one. The definition requires for all $x$ within $\delta$ on either side. If each side separately admits a $\delta$, taking the smaller of the two serves both, so the two-sided limit exists exactly when both one-sided limits exist and agree, with no extra condition.

A limit exists where the function does not. $f(x)={\displaystyle \frac{x^2-4}{x-2}}$ is undefined at $x=2$: substituting gives $0/0$. Yet the limit is 4. Cancelling the common factor leaves $x+2$ for every $x\ne 2$, and the numbers agree, at $x=2.001$ the quotient is $4.001$, at $x=1.999$ it is $3.999$.

Concluding "undefined at the point, so no limit" is the error. The definition excludes $x=a$ deliberately; the value there is irrelevant to where the function is heading.

A value exists where the limit does not. The step function with $\mathrm{sgn}(x)=-1$ for and $+1$ for $x\ge 0$ has $f(0)=1$, perfectly well defined. But the left limit is $-1$ and the right limit is $+1$, so no two-sided limit exists. Having a value is no guarantee of having a limit.

The same shape underlies $\frac{|x|}{x}$, which is $+1$ for and $-1$ for . The disagreement that made $|x|$ non-differentiable at 0.

$0/0$ is not an answer. All three of these are $0/0$ as $x\to 0$:

The form is identical and the answers are a finite nonzero number, zero, and no finite limit at all. Reporting "the limit is $0/0$" states the obstacle, not the result.

The limit laws do not apply where a limit is missing. For $\underset{x\to 0}{lim}{x}^2\sin (1/x)$, the product law would give $0\cdot lim\sin (1/x)$, but $\sin (1/x)$ has no limit at 0, oscillating forever between $-1$ and $1$. The law's hypothesis fails, so the law says nothing.

The limit is nonetheless 0, established by squeezing between $-x^2$ and $x^2$: at $x=0.01$ the function is $-0.0000506$ against a bound of $0.0001$, and at $x=0.001$ it is $0.00000083$ against $0.000001$. A law that does not apply is not evidence that the limit fails to exist.

Tight bounds converging to different values prove nothing. Squeezing requires a common limit. Knowing $-1\le \sin (1/x)\le 1$ near 0 is true and useless: the bounds tend to $-1$ and $1$, the band never narrows, and no conclusion follows.

The intermediate value theorem fails without continuity. The step function above has $f(-1)=-1$ and $f(1)=1$, and takes no value between them anywhere, in particular, never 0. The theorem's conclusion is not almost true here; it is simply false, and continuity was the entire hypothesis.

A removable discontinuity is repairable; a jump is not. Defining $f(2)=4$ makes $\frac{x^2-4}{x-2}$ continuous at 2, because the limit existed and only the value was missing. No assignment at $x=0$ repairs the step function, since no single number equals both $-1$ and $+1$.

Substitution suffices: $\underset{x\to 2}{lim}(x^2-4x)$. Polynomials are continuous everywhere, so the limit is the value: $4-8=-4$. Approaching from either side confirms it, both one-sided computations giving $-4.00000000$. No work beyond evaluation is required, and recognising this case is what keeps the rest of the toolkit for where it is needed.

Algebra first: $\underset{x\to 2}{lim}{\displaystyle \frac{x^2-4}{x-2}}=4$. Substitution gives $0/0$. Factoring the numerator as $(x-2)(x+2)$ and cancelling, legitimate because $x\ne 2$ throughout the approach, leaves $x+2$, whose limit is 4. The numbers agree: $4.001$ at $x=2.001$, $3.999$ at $x=1.999$.

One-sided disagreement: $\underset{x\to 0}{lim}{\displaystyle \frac{|x|}{x}}$. From the right the quotient is $+1$; from the left it is $-1$. Both one-sided limits exist and they differ, so the two-sided limit does not exist. This is a jump discontinuity, and it is the reason $|x|$ has no derivative at the origin.

Squeeze: $\underset{x\to 0}{lim}{x}^2\sin (1/x)=0$. The oscillating factor has no limit, so no law applies. But $-x^2\le x^2\sin (1/x)\le x^2$ for every $x\ne 0$, and both bounds tend to 0. At $x=0.1$ the value is $-0.00544$ inside a bound of $0.01$; at $x=0.001$ it is $0.00000083$ inside $0.000001$. The band collapses and carries the function with it.

At infinity: $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3x^2+2x}{x^2-5}}=3$. Dividing numerator and denominator by $x^2$ gives $\frac{3+2/x}{1-5/x^2}$, and the corrections vanish. The convergence is visible: $3.368$ at $x=10$, $3.0215$ at $x=100$, $3.00202$ at $x=1000$, $3.000002$ at $x={10}^6$. Equal degrees, so the answer is the ratio of leading coefficients.

A limit that is infinite: $\underset{x\to 0}{lim}{\displaystyle \frac{1}{x^2}}=\mathrm{\infty }$. The quotient grows without bound, $100$ at $x=0.1$, ${10}^6$ at $x=0.001$. Writing $\mathrm{\infty }$ records the manner of failure rather than naming a number: no real value is approached, which is why this is an infinite discontinuity and why ${\int }_{-1}^1{x}^{-2}dx$ diverges.

The first is the common case, and the only one needing no technique. Each of the others marks a distinct obstruction, a hole, a jump, an oscillation, an unbounded interval, an unbounded value, and each has its own method. Diagnosing which case is in front of you is most of the skill.

The squeeze theorem. If $g(x)\le f(x)\le h(x)$ for all $x$ near $a$ (except possibly at $a$), and

then $\underset{x\to a}{lim}f(x)=L$.

Proof. Let . Since $g\to L$ there is ${\delta }_1$ with whenever ; since $h\to L$ there is ${\delta }_2$ with on the corresponding interval. Take $\delta =min({\delta }_1,{\delta }_2)$. Then for ,

so . $◼$

The proof shows why a common limit is required: the two inequalities pin $f$ from below and above to the same number, and with different limits the outer bounds would never meet.

Application. $\underset{x\to 0}{lim}{x}^2\sin \left(\frac{1}{x}\right)=0$. Since $-1\le \sin (1/x)\le 1$ for every $x\ne 0$, multiplying by gives $-x^2\le x^2\sin (1/x)\le x^2$, and both bounds tend to 0. Checking: at $x=0.01$ the function is $-0.0000506366$ with bound $0.0001$; at $x=0.0001$ it is $-0.0000000031$ with bound.

No limit law reaches this, because $\sin (1/x)$ has no limit at 0 to feed the product rule.

The intermediate value theorem. If $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then there is $c\in [a,b]$ with $f(c)=N$.

What it guarantees and what it does not. Existence only. It does not say how many such $c$ there are, nor where, nor how to find one. And it is false without continuity: the sign function has $f(-1)=-1$, $f(1)=1$ and never takes the value 0.

Application to root-finding. For $p(x)=x^3-x-2$: $p(1)=-2$ and $p(2)=4$. The polynomial is continuous, and 0 lies between $-2$ and $4$, so a root exists in $(1,2)$.

Bisection turns the theorem into an algorithm: test the midpoint, keep whichever half still shows a sign change, repeat. Each step halves the bracket, so the uncertainty after $n$ steps is $2^{-n}$. Sixty steps on this polynomial give $c\approx 1.5213797068$, with $p(c)$ of order ${10}^{-15}$.

Why continuity is the whole hypothesis. The method assumes that a sign change across an interval implies a crossing inside it. For a function with a jump, the sign can change by leaping over zero rather than passing through it, and every bisection step would preserve a bracket containing no root at all.

1. An indeterminate form cleared by factoring: $\underset{x\to 3}{lim}{\displaystyle \frac{x^2-9}{x^2-x-6}}$.

Substituting gives $\frac{0}{0}$. Both numerator and denominator vanish at 3, so both carry a factor of $(x-3)$:

Now the quotient law applies, since the denominator tends to $5\ne 0$:

2. An indeterminate form cleared by rationalising: $\underset{x\to 0}{lim}{\displaystyle \frac{\sqrt{x+4}-2}{x}}$.

Again $0/0$. Multiply above and below by the conjugate $\sqrt{x+4}+2$:

The limit is $\frac{1}{\sqrt{4}+2}=\frac{1}{4}=0.25$.

Check numerically: at $x=0.0001$ the original expression is.

3. One-sided limits that disagree: at $x=1$.

From the left, $f$ follows $x^2$, so $\underset{x\to 1^{-}}{lim}f(x)=1$. From the right it follows $3x$, so $\underset{x\to 1^{+}}{lim}f(x)=3$.

The one-sided limits exist and differ, so the two-sided limit does not exist, and $f$ has a jump discontinuity at 1 of size 2. The value $f(1)=3$ is defined, which changes nothing. A value is not a limit.

4. A limit at infinity: $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{3x^2+2x}{x^2-5}}$.

Divide numerator and denominator by the highest power present, $x^2$:

Check: the values run $3.3684$ at $x=10$, $3.0215$ at $x=100$, $3.0020$ at $x=1000$, $3.000002$ at $x={10}^6$.

The rule for rational functions follows from this manoeuvre: equal degrees give the ratio of leading coefficients, a smaller numerator degree gives 0, a larger one gives $\pm \mathrm{\infty }$.

5. Continuity decided against all three conditions: $g(x)={\displaystyle \frac{x^2-1}{x-1}}$ at $x=1$, with $g(1)=5$ assigned.

So $g$ is discontinuous at 1, and the discontinuity is removable: the limit exists, so redefining $g(1)=2$ makes it continuous. The numbers confirm the limit, $g(1.001)=2.001$ and $g(0.999)=1.999$.

Had the third condition been the one that held while the second failed, no redefinition would have helped. Checking in order is what makes the classification immediate.