Limits, Continuity and Differentiability

Every differentiable function is continuous. Is every derivative continuous?

No. Take

$f(x)=\{\begin{array}{ll}{x}^2\sin \frac{1}{x},& x\ne 0\\ 0,& x=0.\end{array}$

At the origin the definition gives $f^{\prime }(0)={\lim }_{h\to 0}h\sin \frac{1}{h}=0$, because $\left|h\sin \frac{1}{h}\right|\le |h|$ and the squeeze theorem finishes it. So $f$ is differentiable everywhere. But away from the origin,

$f^{\prime }(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x},$

and as $x\to 0$ the first part tends to $0$ while $\cos \frac{1}{x}$ oscillates between $-1$ and $1$ forever. So ${\lim }_{x\to 0}{f}^{\prime }(x)$ does not exist, even though $f^{\prime }(0)$ does.

This is the shape of the whole chapter at Advanced level. Main tests functions that behave; Advanced tests functions built specifically to separate two conditions that usually travel together — a limit from a value, continuity from differentiability, the existence of a derivative from the continuity of that derivative. Every such example is constructed by attaching an oscillation, a modulus, a greatest-integer function or a piecewise definition at exactly one point.

1. What a limit requires, and the standard ones

A limit exists only when both one-sided limits exist and agree. That is the whole definition, and almost every "does the limit exist" question is answered by computing the two sides separately — which is compulsory whenever the expression contains a modulus, a greatest-integer function, an exponent that changes sign, or a piecewise definition.

The standard limits are worth knowing as a block, because they are the vocabulary in which every harder limit is expressed:

${\lim }_{x\to 0}\frac{\sin x}{x}=1,{\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2},{\lim }_{x\to 0}\frac{e^x-1}{x}=1,{\lim }_{x\to 0}\frac{\ln (1+x)}{x}=1,$

together with ${\lim }_{x\to 0}\frac{a^x-1}{x}=\ln a$ and ${\lim }_{x\to 0}\frac{(1+x{)}^n-1}{x}=n$.

Two further tools handle everything the standard limits do not reach. The squeeze theorem says that if $g(x)\le f(x)\le h(x)$ near $a$ and both outer functions tend to the same value $L$, then so does $f$. It is the only way to evaluate a limit whose expression oscillates, since no algebraic manipulation can tame $\sin \frac{1}{x}$.

The other tool is a conversion. The forms $\infty -\infty$ and $0\cdot \infty$ are not directly indeterminate in any usable sense, so they are rewritten as $\frac{0}{0}$ or $\frac{\infty }{\infty }$ before anything else happens: combine two fractions over a common denominator for the first, and move one factor into a denominator as its reciprocal for the second.

A limit as $x\to \infty$ is handled by dividing numerator and denominator by the highest power present, or by substituting $x=\frac{1}{t}$ and letting $t\to 0$, which converts every result in this chapter into a statement about the origin.

Illustration 1

Evaluate $\underset{x\to 0}{\lim }\frac{\sin x^{\circ }}{x}$.

The first standard limit is a statement about radians. Converting, $x^{\circ }=\frac{\pi x}{180}$ radians, so

$\frac{\sin x^{\circ }}{x}=\frac{\pi }{180}\cdot \frac{\sin \left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}}⟶\frac{\pi }{180}.$

The answer is not $1$. Every trigonometric limit and every derivative of a trigonometric function assumes radian measure, and this is the one place the assumption is tested directly.

Illustration 2

Show that $\underset{x\to 0}{\lim }\frac{|x|}{x}$ does not exist, but $\underset{x\to 0}{\lim }x\left[\frac{1}{x}\right]$ does.

The first has left limit $-1$ and right limit $+1$, so it fails. For the second, the greatest-integer function satisfies $\frac{1}{x}-1<\left[\frac{1}{x}\right]\le \frac{1}{x}$. Multiplying by $x>0$ and squeezing gives a limit of $1$ from the right, and repeating with $x<0$, which reverses the inequalities, also gives $1$. So the limit is $1$.

2. Expansions beat repeated differentiation

For a limit of the form $\frac{0}{0}$ at the origin, substituting the first few terms of a series is faster and safer than applying L'Hopital's rule several times. The four expansions needed are

$\sin x=x-\frac{x^3}{6}+\cdots ,\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots ,e^x=1+x+\frac{x^2}{2}+\cdots ,\ln (1+x)=x-\frac{x^2}{2}+\cdots ,$

along with $\tan x=x+\frac{x^3}{3}+\cdots$. Keep terms only to the order the denominator demands.

Illustration 3

Evaluate $\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}$.

Substituting the expansion, the numerator is $x-\left(x-\frac{x^3}{6}+\cdots \right)=\frac{x^3}{6}+\cdots$, so the limit is $\frac{1}{6}$. Three applications of L'Hopital's rule give the same answer with three chances to slip, and each application requires re-checking that the form is still indeterminate.

Illustration 4

Evaluate $\underset{x\to 0}{\lim }\frac{e^x-1-x-\frac{x^2}{2}}{x^3}$.

The expansion of $e^x$ removes exactly the three subtracted terms, leaving $\frac{x^3}{6}$, so the limit is $\frac{1}{6}$. When a numerator is built to cancel the leading terms of a known series, the expansion is not merely faster — it is the only method that makes the structure visible.

Illustration 5

Show that L'Hopital's rule cannot be used on $\underset{x\to \infty }{\lim }\frac{x+\sin x}{x}$.

The limit itself is easy: $1+\frac{\sin x}{x}\to 1$. But differentiating top and bottom gives $\frac{1+\cos x}{1}$, which oscillates and has no limit at all.

The rule states that if the limit of the ratio of derivatives exists, then it equals the original limit. It never says the converse. Applying it and concluding "the limit does not exist" is a genuine error, and this form is set to catch it.

3. The $1^{\infty }$ form

Expressions such as ${\left(1+\text{small}\right)}^{\text{large}}$ are indeterminate because the base tends to $1$ from an unknown side at an unknown rate. Writing $f^g=e^{g\ln f}$ and expanding $\ln f=\ln \left(1+(f-1)\right)\approx f-1$ gives the working rule

$\lim f^g=\exp \left[\lim g\left(f-1\right)\right],$

valid whenever $f\to 1$ and $g\to \infty$.

Illustration 6

Evaluate $\underset{x\to 0}{\lim }{\left(\cos x\right)}^{1/x^2}$.

Here $f-1=\cos x-1\approx -\frac{x^2}{2}$ and $g=\frac{1}{x^2}$, so the exponent tends to $-\frac{1}{2}$ and the limit is $e^{-1/2}$.

Note what would go wrong without the rule: the base tends to $1$ and the exponent to infinity, and reading the answer as $1$ ignores the rate at which the base approaches $1$.

Illustration 7

Evaluate $\underset{x\to 0}{\lim }{\left(\frac{1+\tan x}{1+\sin x}\right)}^{1/x^3}$.

The base tends to $1$, so use the rule. Now

$f-1=\frac{\tan x-\sin x}{1+\sin x}=\frac{\sin x\left(1-\cos x\right)}{\cos x\left(1+\sin x\right)}\approx \frac{x\cdot \frac{x^2}{2}}{1}=\frac{x^3}{2}.$

Multiplying by $g=x^{-3}$ gives $\frac{1}{2}$, so the limit is $\sqrt{e}$.

A last remark on when the rule is safe. The exponential form $\exp \left[\lim g(f-1)\right]$ relies on $\ln f\approx f-1$, which is only the first term of the logarithmic series, so it is valid precisely because $f\to 1$ makes every later term negligible.

If the base does not tend to $1$ the form is not indeterminate at all: a base tending to a number below $1$ with an exponent tending to $+\infty$ gives $0$, and a base above $1$ gives $\infty$, both directly.

4. Continuity and its three failures

A function is continuous at $a$ when $f(a)$ exists, ${\lim }_{x\to a}f(x)$ exists, and the two are equal. Each of the three can fail separately, and the vocabulary matters because questions ask which kind of discontinuity is present.

Illustration 8

Find $a$ and $b$ so that $f$ is continuous everywhere, where $f(x)=\frac{\sin 2x}{x}$ for $x<0$, $f(0)=a$, and $f(x)=b+x^2$ for $x>0$.

The left limit is $2$, from the standard limit with the factor doubled. The right limit is $b$. Continuity needs both to equal $f(0)=a$, so $a=b=2$.

Every "find the constants" question reduces to this: compute both one-sided limits, then set them equal to the stated value at the point.

5. Differentiability and the two one-sided derivatives

The derivative at $a$ exists when

${\lim }_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h}\text{and}{\lim }_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}$

both exist and are equal. Differentiability implies continuity, because the numerator must tend to zero for a finite quotient to survive; the converse fails at every corner.

Illustration 9

Show that $g(x)=x|x|$ is differentiable at the origin although $|x|$ is not.

For $x\ge 0$, $g(x)=x^2$ with right derivative $0$ at the origin; for $x<0$, $g(x)=-x^2$ with left derivative $0$. The two agree, so $g^{\prime }(0)=0$ and in fact $g^{\prime }(x)=2|x|$ everywhere.

Multiplying by $x$ has smoothed the corner: near the origin the two branches leave with the same slope instead of with slopes $\pm 1$. The general principle is that each extra factor of $x$ buys one more order of smoothness, which is exactly why $x\sin \frac{1}{x}$ is continuous but not differentiable at the origin while $x^2\sin \frac{1}{x}$ is differentiable.

A modulus wrapped around a function follows a rule worth stating on its own. Where $f(a)\ne 0$, the sign of $f$ is fixed nearby, so $|f|$ is just $\pm f$ and is differentiable exactly where $f$ is. Where $f(a)=0$, the graph of $|f|$ is reflected upwards, and the reflection creates a corner unless the graph arrives flat. So

$|f|\text{is differentiable at a zero of }f⟺f^{\prime }(a)=0.$

That is why $|x|$ fails at the origin while $\left|x^2\right|$ does not: the first crosses zero with slope $1$, the second touches zero with slope $0$. The same rule governs a maximum of two functions, since $\max (f,g)=\frac{1}{2}\left(f+g+|f-g|\right)$, so the corners of a maximum sit exactly where the two graphs cross with different slopes.

Illustration 10

Where is $h(x)=|x-1|+|x-2|$ not differentiable, and what is the minimum value of $h$?

Corners occur only where a modulus changes sign, so at $x=1$ and $x=2$. Between them $h(x)=(x-1)+(2-x)=1$, constant; outside them it is $\pm (2x-3)$, with slopes $-2$ and $+2$.

So $h$ is not differentiable at exactly two points, and its minimum value is $1$, attained on the whole interval $[1,2]$. Reading the slopes on each piece answers both parts at once.

Illustration 11

A differentiable function satisfies $f(x+y)=f(x)f(y)$ for all reals, with $f(0)\ne 0$ and $f^{\prime }(0)=2$. Find $f$.

Putting $y=0$ gives $f(0)=1$. From the definition of the derivative,

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}=f(x){\lim }_{h\to 0}\frac{f(h)-1}{h}=2f(x).$

So $\frac{f^{\prime }}{f}=2$, giving $f(x)=C{e}^{2x}$, and $f(0)=1$ fixes $C=1$. The whole solution came from turning the functional equation into a differential equation using nothing but the definition of the derivative.

6. The three theorems, and what they actually promise

The intermediate value theorem says a function continuous on $[a,b]$ takes every value between $f(a)$ and $f(b)$. It guarantees a root but says nothing about how many or where.

Rolle's theorem says that if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c$ inside. The Lagrange mean value theorem drops the equal-endpoint condition and concludes $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$: some tangent is parallel to the chord.

All three conclusions are existence statements. None of them locates the point, and none of them is reversible.

Illustration 12

Show that $x^3-3x+k=0$ has at most one root in $[0,1]$, whatever the value of $k$.

Suppose there were two roots. Rolle's theorem, applied to $p(x)=x^3-3x+k$, would give a point in $(0,1)$ where $p^{\prime }(x)=3x^2-3=0$, that is $x=\pm 1$. Neither lies inside $(0,1)$, so no such point exists and there cannot be two roots.

Rolle's theorem is used this way far more often than for its own sake: assume two roots, derive a stationary point, and show none is available.

Illustration 13

Prove that $\left|\sin a-\sin b\right|\le |a-b|$ for all reals.

Apply the mean value theorem to $\sin$ on the interval between $a$ and $b$: there is a $c$ with $\sin a-\sin b=(a-b)\cos c$. Since $|\cos c|\le 1$, taking moduli gives the result.

Every inequality of the form "the change in $f$ is at most $M$ times the change in $x$" is this theorem with a bound on $f^{\prime }$.

Illustration 14

Show that $x{2}^x=1$ has a solution in $(0,1)$.

Let $\varphi (x)=x{2}^x-1$, which is continuous. Then $\varphi (0)=-1$ and $\varphi (1)=1$, so the intermediate value theorem provides a root strictly between. Since $\varphi$ is increasing on $[0,1]$, that root is unique — but uniqueness came from monotonicity, not from the theorem.

Illustration 15

If $f$ is differentiable on $\mathbb{R}$ with $f^{\prime }(x)=0$ everywhere, prove that $f$ is constant.

Take any $a<b$. The mean value theorem gives a $c$ in $(a,b)$ with $f(b)-f(a)=f^{\prime }(c)(b-a)=0$, so $f(b)=f(a)$. Since $a$ and $b$ were arbitrary, $f$ is constant.

The result looks obvious and is not: it is exactly what the mean value theorem is for, and no argument avoiding the theorem is available at this level.

Summary

A limit exists only if both one-sided limits exist and agree, which is why any modulus, greatest-integer function or piecewise definition forces a two-sided calculation. The standard limits are stated in radians, and a question posed in degrees changes the answer by a factor of $\frac{\pi }{180}$.

For $\frac{0}{0}$ forms at the origin, substitute series expansions and keep terms only to the order the denominator needs. L'Hopital's rule is available but conditional: it concludes only when the limit of the derivative ratio exists, and a failure of that ratio proves nothing about the original limit. The $1^{\infty }$ form is handled by $\lim f^g=\exp \left[\lim g(f-1)\right]$, which encodes the rate at which the base approaches $1$.

Continuity requires the value, the limit, and their equality, and the failures are named: removable, jump, infinite and oscillatory. Differentiability requires the two one-sided derivatives to exist and agree; it implies continuity and is not implied by it. A derivative need not be continuous, as $x^2\sin \frac{1}{x}$ shows, and each extra factor of $x$ buys one more order of smoothness at the origin.

A modulus placed around a function is differentiable at a zero only if the function arrives there flat, which is why $|x|$ fails and $x|x|$ does not, and why the corners of a maximum of two graphs sit exactly at their transversal crossings.

The three theorems are existence statements. The intermediate value theorem produces a root, Rolle's theorem produces a stationary point between equal values, and the mean value theorem produces a tangent parallel to the chord. None locates its point, and the standard uses are indirect: bounding a difference, or showing that a second root would force an impossible stationary point.