Limits, Continuity and Differentiability

${\lim }_{x\to 0}\frac{\sin x}{x}=1$

Everyone knows that one. Now two expressions that look almost identical.

The second fails because the modulus treats the two sides differently. Approaching from the right, $|x|=x$ and the ratio tends to $+1$. From the left, $|x|=-x$ and it tends to $-1$. Two answers means no answer.

The third fails because $\sin x^{\circ }$ means $\sin \left(\frac{\pi x}{180}\right)$, so the standard limit returns $\pi /180$, a number fifty-seven times smaller than $1$. Every standard limit in this chapter assumes radians.

Do the two sides agree, and about what? That single question, asked three times with rising strictness, is the entire chapter.

Each condition contains the one before it, which is why differentiability implies continuity and continuity implies a limit, and why neither arrow reverses.

1. Functions, Families and Graphs

Before limits, know the shapes. Most limit questions are answered by recognising the family rather than by manipulation.

The last column is where discontinuities come from. If you can name the family, you can predict where the trouble will be before doing any algebra.

Illustration 1

On what set can $f(x)=\frac{\ln (x-1)}{\sqrt{4-x^2}}$ possibly be continuous?

Continuity can only be discussed where the function is defined, so find the domain first. Each ingredient imposes its own requirement.

$\ln (x-1)\Rightarrow x>1,\sqrt{4-x^2}\text{ in a denominator}\Rightarrow 4-x^2>0\Rightarrow -2<x<2$

Both requirements hold at once, so intersect them.

$\text{domain}=(1,2)$

On that open interval the function is a quotient of continuous functions with a non-vanishing denominator, so it is continuous throughout, and there is nothing left to check.

The lesson is worth generalising: a function built from continuous pieces by sums, products, quotients and composition is continuous wherever it is defined. Discontinuities appear exactly where the definition breaks down, which is why finding the domain does most of the work.

2. Limits: The Two Sides Agreeing

${\lim }_{x\to a}f(x)=L⟺{\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L$

The value $f(a)$ is irrelevant to the limit. A limit describes the approach, not the arrival, which is why a function with a hole at $a$ can still have a limit there.

Trap. Wherever a modulus, a greatest-integer function or a piecewise definition appears, check the two sides separately. They are the only situations where the two sides can genuinely differ.

Illustration 2

Evaluate $\underset{x\to 0}{\lim }\frac{\sin x}{|x|}$, and explain why the modulus changes the answer.

Split at $0$, because that is exactly where $|x|$ changes its formula.

$x>0:\frac{\sin x}{|x|}=\frac{\sin x}{x}⟶1$

$x<0:\frac{\sin x}{|x|}=\frac{\sin x}{-x}=-\frac{\sin x}{x}⟶-1$

The two sides give $+1$ and $-1$, so the limit does not exist.

Notice what happened: the numerator is odd and the denominator became even, so the ratio changed sign across zero. Without the modulus both are odd and the ratio is even, which is why $\sin x/x$ approaches the same value from both sides.

3. Evaluating Limits

Substitute first. If the result is a number, that is the limit. Only if it is indeterminate does work begin.

Trap. $\frac{1}{0}$ and $\frac{0}{5}$ are not indeterminate. The first is unbounded and the second is zero, and neither needs any technique.

The standard limits, all in radians, do most of the work.

${\lim }_{x\to 0}\frac{\sin x}{x}=1,{\lim }_{x\to 0}\frac{\tan x}{x}=1,{\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$

${\lim }_{x\to 0}\frac{a^x-1}{x}=\ln a,{\lim }_{x\to 0}\frac{\log (1+x)}{x}=1,{\lim }_{x\to \infty }{\left(1+\frac{k}{x}\right)}^x=e^k$

Illustration 3

Evaluate $\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$.

Substituting gives $0/0$. Two surds separated by a minus sign is the signal to rationalise.

$\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\cdot \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{(1+x)-(1-x)}{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}$

$=\frac{2x}{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}$

The offending $x$ cancelled, which is the whole purpose of the manoeuvre, and substitution is now safe.

$⟶\frac{2}{1+1}=1$

Illustration 4

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

There is no standard limit for this shape, but there is one for $\frac{a^x-1}{x}$. Manufacture it by subtracting and adding $1$.

$\frac{3^x-2^x}{x}=\frac{\left(3^x-1\right)-\left(2^x-1\right)}{x}=\frac{3^x-1}{x}-\frac{2^x-1}{x}$

Both pieces are now standard, and the algebra of limits allows the split because each piece has a finite limit.

$⟶\ln 3-\ln 2=\ln \frac{3}{2}$

Inserting a $1$ that cancels is one of the two most useful moves in limit questions; rationalising is the other.

Illustration 5

Evaluate $\underset{x\to \infty }{\lim }{\left(1+\frac{3}{x}\right)}^{2x}$.

The base tends to $1$ and the exponent to infinity, which is the indeterminate form $1^{\infty }$. Reshape it into the standard limit.

${\left(1+\frac{3}{x}\right)}^{2x}={\left[{\left(1+\frac{3}{x}\right)}^x\right]}^2⟶{\left(e^3\right)}^2=e^6$

Trap. $1^{\infty }$ is indeterminate, not $1$. The base is never exactly $1$; it is approaching $1$ while being raised to a power growing without bound, and the two effects compete.

Illustration 6

Evaluate $\underset{x\to 0}{\lim }{x}^2\sin \frac{1}{x}$.

The sine has no limit at all as $x\to 0$: it oscillates between $-1$ and $1$ infinitely often. So the product rule for limits does not apply.

Bound it instead.

$-1\le \sin \frac{1}{x}\le 1\Rightarrow -x^2\le x^2\sin \frac{1}{x}\le x^2$

Both outer bounds tend to $0$, so the middle is trapped.

${\lim }_{x\to 0}{x}^2\sin \frac{1}{x}=0$

This is the sandwich theorem, and it is the tool for any limit containing a bounded but wildly behaved factor. Note that $x\sin (1/x)$ works the same way, while $\sin (1/x)$ alone has no limit, since nothing shrinks it.

Algebra of limits

Limits of sums, products and quotients split into limits of the parts, provided each part has a limit and no denominator tends to zero. Illustration 5 shows what happens when that proviso fails.

Limits at infinity

Divide numerator and denominator by the highest power present, then read off which terms survive.

Illustration 7

Evaluate $\underset{x\to \infty }{\lim }\left(\sqrt{x^2+x}-x\right)$.

Substituting gives $\infty -\infty$. Rationalise, treating the expression as a difference over $1$.

$\left(\sqrt{x^2+x}-x\right)\cdot \frac{\sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}=\frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\frac{x}{\sqrt{x^2+x}+x}$

Now divide top and bottom by $x$, the highest power present.

$=\frac{1}{\sqrt{1+\frac{1}{x}}+1}⟶\frac{1}{1+1}=\frac{1}{2}$

The answer is finite, which the original form gave no hint of. A guess of $0$ or $\infty$ would have been equally plausible and both are wrong.

4. Continuity: The Value Attained as Well

$f\text{ continuous at }a⟺{\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=f(a)$

Three things must coincide: both one-sided limits and the actual value. Failing any one of them is a discontinuity.

Illustration 8

Classify the discontinuity of each function at the stated point.

$\text{(i)}\frac{x^2-1}{x-1}\text{ at }x=1,\text{(ii)}\frac{1}{x-1}\text{ at }x=1,\text{(iii)}\lfloor x\rfloor \text{ at }x=2$

(i) Factor: for $x\ne 1$ the function equals $x+1$, so both sides approach $2$. The value at $1$ is simply undefined, and defining $f(1)=2$ repairs it. Removable.

(ii) As $x\to 1^{-}$ the function tends to $-\infty$, and as $x\to 1^{+}$ to $+\infty$. Nothing can be assigned at $1$ to fix this. Infinite.

(iii) From the left the greatest integer is $1$; from the right it is $2$. Two different finite values. Jump.

Only the first is repairable, and that is what "removable" records. The classification also tells you what a question can ask: removable discontinuities produce "find $k$ so that $f$ is continuous", and jumps never do.

5. Differentiability: The Slopes Agreeing

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

The function is differentiable at $a$ when the left-hand and right-hand versions of that limit both exist and agree.

$L{f}^{\prime }(a)={\lim }_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h},R{f}^{\prime }(a)={\lim }_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}$

Three distinct ways to be continuous and still fail: the slopes may disagree finitely (a corner), or run off in opposite directions (a cusp), or both run off the same way (a vertical tangent). All three are continuous, and none is differentiable.

Illustration 9

Is $f(x)=x^{2/3}$ differentiable at $x=0$?

It is certainly continuous there, since $f(0)=0$ and $x^{2/3}\to 0$ from both sides.

$f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}=\frac{2}{3\sqrt[3]{x}}$

$x\to 0^{+}:f^{\prime }\to +\infty ,x\to 0^{-}:f^{\prime }\to -\infty$

Neither one-sided derivative exists as a finite number, and they head in opposite directions. Not differentiable, and the graph has a cusp: a sharp point with two vertical half-tangents.

Compare $x^{1/3}$, where $f^{\prime }(x)=\frac{1}{3}{x}^{-2/3}$ tends to $+\infty$ from both sides. That is a vertical tangent rather than a cusp, and it is still not differentiable, because a derivative must be a finite number.

6. The Implication Chain

$\text{differentiable at }a\Rightarrow \text{continuous at }a\Rightarrow \text{limit exists at }a$

Neither arrow reverses, and each reverse failure has a standard witness.

Trap. The chain is quoted backwards more often than any other fact in this unit. Continuity is the weaker condition; differentiability is the stronger one and therefore the one that implies the other.

Illustration 10

Let $f(x)=x^2\sin \frac{1}{x}$ for $x\ne 0$ and $f(0)=0$. Is $f$ differentiable at $0$? Is $f^{\prime }$ continuous there?

Differentiating the formula and substituting $0$ is not available, since the formula does not apply at $0$. Use the definition.

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

by the sandwich theorem, since the sine is bounded. So $f$ is differentiable at $0$, with $f^{\prime }(0)=0$.

Now differentiate away from $0$ using the product and chain rules.

$f^{\prime }(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x}(x\ne 0)$

As $x\to 0$ the first term vanishes but $\cos (1/x)$ oscillates between $-1$ and $1$ for ever, so $f^{\prime }$ has no limit at $0$.

$f\text{ differentiable everywhere},f^{\prime }\text{ not continuous at }0$

Differentiability of $f$ says nothing about continuity of $f^{\prime }$. The chain of implications runs between the levels at a point, not between a function and its derivative.

7. Rules of Differentiation

Trap. A function built from continuous pieces is continuous wherever it is defined, so finding the domain does most of the work.

Differentiability of $f$ says nothing about continuity of $f^{\prime }$: $x^2\sin (1/x)$ is differentiable everywhere and its derivative is discontinuous at $0$.

The quotient rule's numerator is $u^{\prime }v-u{v}^{\prime }$, and reversing it changes the sign of the whole answer. Remember it as "derivative of the top times the bottom, minus the top times the derivative of the bottom".

Illustration 11

Differentiate $y={\sin }^3(2x+1)$.

Three layers, so the chain rule applies twice. Work from the outside inwards, differentiating one layer at a time and multiplying.

$\frac{dy}{dx}=3{\sin }^2(2x+1)\cdot \cos (2x+1)\cdot 2=6{\sin }^2(2x+1)\cos (2x+1)$

The factor of $2$ at the end is the derivative of the innermost layer, and dropping it is the standard error. A check: at $x$ where $\sin (2x+1)=0$ the derivative should vanish, and it does.

8. The Standard Derivatives

Implicit differentiation

When $y$ is not isolated, differentiate every term with respect to $x$, attaching $dy/dx$ each time a $y$ is differentiated, then solve for $dy/dx$.

Illustration 12

Find the slope of $x^3+y^3=6xy$ at the point $(3,3)$.

Solving for $y$ is impossible here, which is exactly when implicit differentiation earns its place. The right side needs the product rule.

$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$

Collect the derivative terms on one side and factor.

$\frac{dy}{dx}\left(3y^2-6x\right)=6y-3x^2\Rightarrow \frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}$

$\text{at }(3,3):\frac{6-9}{9-6}=-1$

Confirm the point is on the curve first, as an implicit answer is meaningless otherwise: $27+27=54$ and $6(3)(3)=54$.

Logarithmic differentiation

Take logarithms first whenever the expression is a long product or quotient, or has a variable in the exponent. Logarithms turn products into sums, which the sum rule then handles.

Illustration 13

Differentiate $y=\frac{(x-1{)}^2\sqrt{x+2}}{(x+3{)}^3}$.

The quotient and product rules together would take half a page. Take logarithms and the structure collapses into three separate terms.

$\ln y=2\ln (x-1)+\frac{1}{2}\ln (x+2)-3\ln (x+3)$

Differentiate both sides, remembering that the left gives $\frac{1}{y}\frac{dy}{dx}$ by the chain rule.

$\frac{1}{y}\frac{dy}{dx}=\frac{2}{x-1}+\frac{1}{2(x+2)}-\frac{3}{x+3}$

$\frac{dy}{dx}=\frac{(x-1{)}^2\sqrt{x+2}}{(x+3{)}^3}\left[\frac{2}{x-1}+\frac{1}{2(x+2)}-\frac{3}{x+3}\right]$

Every exponent has become a coefficient, which is the whole gain. The method also handles $x^{\sin x}$ and similar, where no other rule applies at all.

Second derivatives

Differentiate the first derivative again. For implicit or parametric functions, remember that the second differentiation must be carried out by the same rules as the first, so a $dy/dx$ appearing inside must itself be differentiated.

Summary

One question, asked three times: do the two sides agree, and about what?

A limit needs the two sides to agree about the value approached; $f(a)$ is irrelevant to it.

Check the two sides separately wherever a modulus, greatest-integer function or piecewise definition appears. $\sin x/|x|$ has no limit at $0$ for exactly that reason.

Every standard limit assumes radians: in degrees, $\sin x^{\circ }/x$ tends to $\pi /180$, not $1$.

Substitute first, and only work if the result is indeterminate. $1/0$ and $0/5$ are not indeterminate.

$0/0$ yields to factoring, rationalising, or inserting a $1$ that cancels; $\infty /\infty$ to dividing by the highest power; $\infty -\infty$ to rationalising; $1^{\infty }$ to the exponential standard limit.

The sandwich theorem handles any bounded but oscillating factor, which is why $x^2\sin (1/x)\to 0$ while $\sin (1/x)$ has no limit.

Continuity needs both one-sided limits and the actual value to coincide. Discontinuities are removable, jump or infinite, and only the first can be repaired.

Differentiability needs the two one-sided derivatives to exist and agree. Corners, cusps and vertical tangents are three distinct ways to be continuous and still fail.

Differentiable implies continuous implies a limit exists, and no arrow reverses. $|x|$ and a factorable hole are the standard witnesses.

The quotient rule's numerator is $u^{\prime }v-u{v}^{\prime }$, and the chain rule multiplies one factor per layer, including the innermost.

Implicit differentiation attaches $dy/dx$ whenever a $y$ is differentiated; check the point lies on the curve before quoting a slope.

Logarithmic differentiation turns long products, quotients and variable exponents into sums, and is the only route for expressions like $x^{\sin x}$.