Limits

The secant slopes of the previous lesson closed in on a number without ever reaching it, and that behaviour needs a definition before anything can be built on it.

The word usually offered is "approaches", which is a description rather than a definition: it says what the numbers are doing without saying what would settle whether a claimed answer is right. This lesson replaces it with something checkable, and then, having made limits precise, shows that most of them are computed in one step by substitution. The interesting ones are exactly the ones where substitution fails.

The tolerance game

Consider the claim that $f (x) = 2 + x$ approaches $2$ as $x$ approaches $0$ . It is obviously true, but ask what makes it true and the answer is not "the values get near $2$ ", since the values also get near $2.0001$ and near $3$ .

The claim that distinguishes $2$ from every other candidate is this: name any tolerance, however small, and I can keep $f (x)$ within it. Ask for $f (x)$ within $0.01$ of $2$ and I keep $x$ within $0.01$ of $0$ . Ask for $0.000001$ and I keep $x$ within $0.000001$ . No other number survives this game. If you propose $2.0001$ as the limit and I ask for a tolerance of $0.00001$ , you cannot deliver, because once $x$ is small the values are nowhere near $2.0001$ .

The limit is therefore not a description of motion but a guarantee about tolerances, and the guarantee has two players: a challenge, and a response that must exist for every challenge. Weierstrass turned exactly this into notation in the 1860s. Writing $ε$ for the challenge tolerance on the output and $δ$ for the response tolerance on the input,

lim_{x \to a} f (x) = L

means: for every $ε > 0$ there is a $δ > 0$ such that $0 < | x - a | < δ$ forces $| f (x) - L | < ε$ .

Two details in that sentence carry real weight. The order is fixed: $ε$ is named first and $δ$ may depend on it, never the reverse. And the condition on the input is $0 < | x - a |$ , which explicitly excludes $x = a$ . The limit is a statement about the punctured neighbourhood of $a$ , and it is deliberately blind to the value of $f$ at $a$ , which may be anything or may not exist. That blindness is not a technicality: it is the only reason the definition is usable for difference quotients, which are always undefined at the point of interest.

Producing a delta

The definition earns its keep when a $δ$ is actually produced, so here is one. Claim: $lim_{x \to 3} (4 x - 5) = 7$ . Given $ε$ , the requirement is $| (4 x - 5) - 7 | < ε$ , which is $| 4 x - 12 | < ε$ , which is $4 | x - 3 | < ε$ . So taking $δ = ε / 4$ works, for every $ε$ at once. The proof is the formula for $δ$ .

A curve needs one extra move. Claim: $lim_{x \to 2} x^{2} = 4$ . The quantity to control is $| x^{2} - 4 | = | x - 2 | \cdot | x + 2 |$ . The first factor is what $δ$ controls directly; the second must be bounded before it can be used. So agree in advance that $δ$ will never exceed $1$ , which confines $x$ to the interval from $1$ to $3$ and makes $| x + 2 | < 5$ . Then $| x^{2} - 4 | < 5 | x - 2 | < 5 δ$ , so choosing $δ$ to be whichever of $1$ and $ε / 5$ is smaller finishes it.

Check it numerically with $ε = 0.05$ , which gives $δ = 0.01$ . At $x = 2.01$ the error is $| 4.0401 - 4 | = 0.0401$ , and at $x = 1.99$ it is $0.0399$ . Both are under the requested $0.05$ , with room to spare, which is normal: $δ$ has to work, not to be optimal.

Nobody computes everyday limits this way, any more than anyone adds by returning to the axioms of arithmetic. The definition is there so that the shortcuts can be proved, and so that a disputed limit has a court of appeal.

The limit laws, and why substitution usually works

From the definition it can be proved, once and for all, that limits respect arithmetic. If $lim_{x \to a} f (x) = L$ and $lim_{x \to a} g (x) = M$ , then the limit of $f + g$ is $L + M$ , of $f - g$ is $L - M$ , of $f g$ is $L M$ , and of $f / g$ is $L / M$ provided $M \neq 0$ . Constants pull out, and powers and roots behave as expected.

Add the two limits that are true directly from the definition, $lim_{x \to a} c = c$ and $lim_{x \to a} x = a$ , and a large class of limits collapses. A polynomial is built from constants and $x$ by multiplication and addition, so for any polynomial $P$ , $lim_{x \to a} P (x) = P (a)$ : substitute and stop. A rational function is a quotient of polynomials, so the same holds wherever the denominator is not zero at $a$ .

That is the useful summary. For every function assembled by arithmetic from polynomials, the limit is the value, and the calculation is one substitution. So $lim_{x \to 2} (x^{3} - 4 x + 1) = 8 - 8 + 1 = 1$ , and there is nothing more to say about it.

The whole subject therefore lives in the exceptions, and there is essentially one: the denominator goes to zero at the same moment as the numerator. The law for quotients does not apply, the expression $0 / 0$ is not defined, and yet the limit frequently exists. Such an expression is called an indeterminate form, and the name is precise. It is not that the answer is unknown; it is that the form itself determines nothing, since $0 / 0$ arises in limits equal to $6$ , to $0$ , to $- 1 / 4$ and to nothing at all.

Working the indeterminate forms

The technique is always the same: change the expression algebraically so that the cancellation happens before the limit is taken. This is legitimate because the limit ignores the point itself, so any manipulation valid for $x \neq a$ is valid here even if it is invalid at $a$ .

For $lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$ , substitution gives $0 / 0$ . Factor: $x^{2} - 9 = (x - 3) (x + 3)$ , so for every $x \neq 3$ the expression equals $x + 3$ , and the limit is $6$ . The cancelled factor is the whole difficulty, and cancelling it is legal for exactly the values the limit is about.

Example. Evaluate $lim_{x \to 2} \frac{x^{3} - 8}{x - 2}$ .

Substitution gives $0 / 0$ , so factor the numerator: $x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)$ , which the factor theorem from the previous course guarantees since $x = 2$ is a root. For $x \neq 2$ the quotient is $x^{2} + 2 x + 4$ , a polynomial, so the limit is $4 + 4 + 4 = 12$ . Numerically, at $x = 2.001$ the original expression is $12.006$ , which agrees.

Now you. Evaluate $lim_{x \to 1} \frac{x^{3} - 1}{x^{2} - 1}$ .

Answer

Both parts vanish at $x = 1$ . Factoring gives $(x - 1) (x^{2} + x + 1)$ over $(x - 1) (x + 1)$ , so for $x \neq 1$ the expression is $(x^{2} + x + 1) / (x + 1)$ , and the limit is $3 / 2$ .

Roots need a different device: multiply by the conjugate, which moves the difficulty from the numerator to the denominator where it does no harm.

Example. Evaluate $lim_{h \to 0} \frac{\sqrt{9 + h} - 3}{h}$ .

Multiply top and bottom by $\sqrt{9 + h} + 3$ . The numerator becomes $(9 + h) - 9 = h$ , so the expression is $h / [h (\sqrt{9 + h} + 3)]$ , which for $h \neq 0$ is $1 / (\sqrt{9 + h} + 3)$ . Now substitution is legal and gives $1 / 6 = 0.1667$ . Checking with $h = 0.001$ in the original gives $0.166662$ , so the algebra and the arithmetic agree.

Now you. Evaluate $lim_{h \to 0} \frac{\sqrt{4 + h} - 2}{h}$ .

Answer

The same conjugate trick gives $1 / (\sqrt{4 + h} + 2)$ , so the limit is $1 / 4$ . At $h = 0.001$ the original expression is $0.249984$ .

Both of those are difference quotients in disguise, which is the point. The limit machinery was invented for them.

When there is no limit

A limit can fail to exist, and the ways it fails are worth knowing by sight, because a rule applied to a limit that does not exist produces confident nonsense.

The commonest failure is a jump: the two sides disagree. For $f (x) = | x | / x$ , every positive $x$ gives $+ 1$ and every negative $x$ gives $- 1$ . Approaching zero from the right the values sit at $1$ , from the left at $- 1$ , and no single number is being approached. This motivates one-sided limits, written $lim_{x \to a^{+}}$ and $lim_{x \to a^{-}}$ , with the obvious definitions restricted to one side. The two-sided limit exists exactly when both one-sided limits exist and are equal, which is a theorem, and it is the standard way to test a function defined by different formulas on either side of a point.

The second failure is blowing up. For $f (x) = 1 / x^{2}$ near zero the values exceed any bound you name. Writing $lim_{x \to 0} 1 / x^{2} = \infty$ is standard and convenient, but it does not assert that the limit exists: it is shorthand for a specific mode of not existing, namely that the values eventually exceed every fixed number. Treating that $\infty$ as a quantity to cancel or divide by is how people prove that $1 = 2$ .

The third failure is oscillation. As $x$ approaches zero, $\sin (1 / x)$ runs through every value between $- 1$ and $1$ infinitely often, no matter how small an interval around zero you take. Nothing is approached, nothing blows up, and no tolerance smaller than $1$ can ever be met.

Limits at infinity, and squeezing

Reversing the roles of the two tolerances gives the other kind of limit. Saying $lim_{x \to \infty} f (x) = L$ means: name any $ε$ , and there is a value of $x$ beyond which $f$ stays within $ε$ of $L$ . The graph of such a function has a horizontal asymptote at height $L$ .

For rational functions these limits are settled by dividing top and bottom by the highest power present. Take $(3 x^{2} + 2 x - 1) / (5 x^{2} - x + 4)$ and divide through by $x^{2}$ to get $(3 + 2 / x - 1 / x^{2}) / (5 - 1 / x + 4 / x^{2})$ . Every term with an $x$ downstairs vanishes, so the limit is $3 / 5$ . The arithmetic bears this out: at $x = 10$ the value is $0.6457$ , at $x = 100$ it is $0.60514$ , and at $x = 1000$ it is $0.60052$ .

Example. Evaluate $lim_{x \to \infty} \frac{2 x^{3} - x}{6 x^{3} + 5}$ .

Divide by $x^{3}$ : the expression becomes $(2 - 1 / x^{2}) / (6 + 5 / x^{3})$ , and the limit is $2 / 6 = 1 / 3$ . At $x = 100$ the original is $0.333316$ , closing on $0.3333$ from below.

Now you. Evaluate $lim_{x \to \infty} \frac{4 x^{2} + 7}{x^{3} - 2}$ .

Answer

Divide by $x^{3}$ to get $(4 / x + 7 / x^{3}) / (1 - 2 / x^{3})$ . The numerator goes to zero and the denominator to $1$ , so the limit is $0$ . Whenever the denominator has the higher degree, the limit at infinity is zero.

One further tool is needed later and is easiest to state now. The squeeze theorem says that if $g (x) \leq f (x) \leq h (x)$ near $a$ , and $g$ and $h$ both have limit $L$ at $a$ , then $f$ does too: a function trapped between two things converging to the same place has nowhere else to go.

Its standard use is $f (x) = x^{2} \sin (1 / x)$ near zero. The oscillating factor never leaves $[- 1, 1]$ , so $- x^{2} \leq f (x) \leq x^{2}$ , and both bounds go to zero. Hence $f (x) \to 0$ , despite oscillating infinitely often on the way. The values confirm it: at $x = 0.01$ the function is about $- 5.06 \times 10^{- 5}$ , and at $x = 0.001$ about $8.27 \times 10^{- 7}$ . This theorem is what will deliver the derivative of the sine function, where no algebraic trick works at all.

What has been bought

The limit is now a definition rather than a gesture, and with the limit laws most limits are one substitution. The exceptions are the indeterminate forms, and the two shown here, factoring and rationalising, handle a large fraction of what the next lessons produce.

Notice what has been carefully avoided throughout: any claim about $f (a)$ . The limit at $a$ can exist while $f (a)$ is undefined, as in every difference quotient, or while $f (a)$ is defined and equal to something else entirely, if the function is given a stray value at one point. Functions where nothing so perverse happens, where the limit is the value, are the ones on which the rest of the subject is built, and they have a name and a set of theorems of their own.