Continuity and the Intermediate Value Theorem

Condition 1: is $f(3)$ defined? Yes. The second line of the definition gives $f(3)=5$.

Condition 2: does $\lim_{x\to 3}f(x)$ exist? For $x\ne 3$ we may cancel, because $x-3\ne 0$:

$\frac{x^{2}-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3 \qquad (x\ne 3).$

A limit as $x\to 3$ only uses values with $x\ne 3$, so

$\lim_{x\to 3}f(x)=\lim_{x\to 3}(x+3)=6.$

Both one-sided limits equal $6$, so the limit exists and is finite.

Condition 3: are the two numbers equal? $\lim_{x\to 3}f(x)=6$ but $f(3)=5$, and $6\ne 5$. Condition 3 fails.

Conclusion. $f$ is not continuous at $3$. Because the limit exists and is finite while the assigned value is the wrong one, this is a removable discontinuity. It is repaired by a single redefinition: setting $f(3)=6$, which amounts to using $g(x)=x+3$ for all $x$, makes the function continuous at $3$ and hence continuous on all of $(-\infty,\infty)$.

Factor.

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

The denominator vanishes at $x=1$ and $x=2$, so the domain is all $x$ with $x\ne 1$ and $x\ne 2$. Everywhere else $f$ is a rational function with nonzero denominator, hence continuous by the quotient rule. So the only candidates are $x=1$ and $x=2$.

At $x=1$. For $x\ne 1$ we may cancel the factor $x-1$:

$f(x)=\frac{x+1}{x-2}\qquad (x\ne 1 \text{ and } x\ne 2).$

Therefore

$\lim_{x\to 1}f(x)=\frac{1+1}{1-2}=\frac{2}{-1}=-2,$

a finite limit, while $f(1)$ is undefined (condition 1 fails). This is a removable discontinuity; defining $f(1)=-2$ repairs it.

At $x=2$. Using the same cancelled form, as $x\to 2$ the numerator tends to $3\ne 0$ while the denominator tends to $0$. Checking signs:

• as $x\to 2^{-}$: $x-2\to 0^{-}$, so $\dfrac{x+1}{x-2}\to -\infty$;
• as $x\to 2^{+}$: $x-2\to 0^{+}$, so $\dfrac{x+1}{x-2}\to +\infty$.

(Numerical check: $f(1.999)\approx -2999$ and $f(2.001)\approx 3001$.) This is an infinite discontinuity, with the vertical asymptote $x=2$. It cannot be removed by redefining a single value.

Conclusion. $f$ has a removable discontinuity at $x=1$ and an infinite discontinuity at $x=2$, and it is continuous on

$(-\infty,1)\cup(1,2)\cup(2,\infty).$

Remove the absolute value. By definition $|x-3|=x-3$ when $x>3$, and $|x-3|=-(x-3)$ when $x<3$. Hence for $x\ne 3$,

$f(x)=\begin{cases} \dfrac{-(x-3)}{x-3}=-1, & x<3\\ \dfrac{x-3}{x-3}=+1, & x>3 \end{cases}$

One-sided limits. $f$ is constantly $-1$ just to the left of $3$ and constantly $+1$ just to the right, so

$\lim_{x\to 3^{-}}f(x)=-1, \qquad \lim_{x\to 3^{+}}f(x)=+1.$

One-sided continuity. The definition gives $f(3)=1$.

• Right-continuity: $\lim_{x\to 3^{+}}f(x)=1=f(3)$, so yes, $f$ is right-continuous at $3$.
• Left-continuity: $\lim_{x\to 3^{-}}f(x)=-1\ne 1=f(3)$, so no, $f$ is not left-continuous at $3$.

Continuity. Continuity at $3$ requires both one-sided conditions, so $f$ is not continuous at $3$. Equivalently, the two-sided limit does not exist, so condition 2 of the definition fails.

Classification. Both one-sided limits exist and are finite but are unequal, so this is a jump discontinuity, with jump size

$\lim_{x\to 3^{+}}f(x)-\lim_{x\to 3^{-}}f(x)=1-(-1)=2.$

No choice of $f(3)$ can repair a jump: whatever value is assigned, it can match at most one of the two one-sided limits. Away from $3$ the function is locally constant, so $f$ is continuous on $(-\infty,3)\cup(3,\infty)$.

Left-hand limit. For $x<2$ we certainly have $x\ne 2$, so we may cancel:

$\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2.$

Hence

$\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{-}}(x+2)=4.$

Value and right-hand limit. The second line applies at $x=2$ and to its right, and $ax+1$ is a polynomial, so

$f(2)=2a+1, \qquad \lim_{x\to 2^{+}}f(x)=\lim_{x\to 2^{+}}(ax+1)=2a+1.$

Impose continuity. All three of $\lim_{x\to 2^{-}}f(x)$, $\lim_{x\to 2^{+}}f(x)$ and $f(2)$ must coincide, so

$2a+1=4 \quad\Rightarrow\quad 2a=3 \quad\Rightarrow\quad a=\frac{3}{2}.$

Check. With $a=\dfrac{3}{2}$ we get $f(2)=2\cdot\dfrac{3}{2}+1=4$, which matches the left-hand limit $4$. Continuity holds.

Where is $f$ continuous? For $x<2$, $f$ agrees with the polynomial $x+2$ (the cancellation is valid there), so it is continuous on $(-\infty,2)$. For $x>2$ it is the polynomial $\dfrac{3}{2}x+1$, also continuous. With $a=\dfrac{3}{2}$ the switch point is fine as well, so $f$ is continuous on all of $(-\infty,\infty)$. For any other value of $a$, $f$ has a jump discontinuity at $x=2$ of size $2a+1-4=2a-3\ne 0$.

Step 1: the numerator. Let $u(x)=4-x^{2}$, a polynomial, hence continuous on $(-\infty,\infty)$, and recall that $\ln$ is continuous on $(0,\infty)$. By the composition theorem (if $u$ is continuous at $x$ and $\ln$ is continuous at $u(x)$, then $\ln(u(x))$ is continuous at $x$), the numerator is continuous at every $x$ for which

$4-x^{2}>0, \quad\text{that is}\quad x^{2}<4, \quad\text{that is}\quad -2<x<2.$

Outside $[-2,2]$ the numerator is undefined, and at $x=\pm 2$ the argument of $\ln$ equals $0$, which is also outside the domain of $\ln$.

Step 2: the denominator. $x+1$ is a polynomial, continuous everywhere, and it vanishes only at $x=-1$.

Step 3: the quotient. By the quotient rule for continuity, a quotient of functions continuous at $x$ is continuous at $x$ provided the denominator is nonzero there. Combining Steps 1 and 2, $g$ is continuous at every $x$ with $-2<x<2$ and $x\ne -1$, that is, on

$(-2,-1)\cup(-1,2).$

This is also exactly the domain of $g$, so $g$ is continuous at every point of its domain — as any elementary function is.

Behaviour at $x=-1$. There the numerator tends to $\ln\left(4-(-1)^{2}\right)=\ln 3\approx 1.0986\ne 0$ while the denominator tends to $0$:

• as $x\to -1^{-}$: $x+1\to 0^{-}$, so $g(x)\to -\infty$;
• as $x\to -1^{+}$: $x+1\to 0^{+}$, so $g(x)\to +\infty$.

So $x=-1$ is an infinite discontinuity with vertical asymptote $x=-1$, and it is not removable.

A remark on $x=\pm 2$. These are not classified as discontinuities, because $g$ is undefined on a whole one-sided neighbourhood there ($g$ does not exist for $|x|\ge 2$); they are domain endpoints. For the record, $\ln(4-x^{2})\to -\infty$ as $x\to 2^{-}$ while $x+1\to 3$, so $g(x)\to -\infty$; and as $x\to -2^{+}$ the denominator tends to $-1$, so $g(x)\to +\infty$.

Where can continuity fail? On $(-\infty,-1)$, on $(-1,2)$ and on $(2,\infty)$ the function agrees with a polynomial, and polynomials are continuous. So the only possible discontinuities are the switch points $x=-1$ and $x=2$.

Condition at $x=-1$. The middle line covers $x=-1$, so

$f(-1)=a(-1)+b=-a+b, \qquad \lim_{x\to -1^{+}}f(x)=-a+b,$

while from the left

$\lim_{x\to -1^{-}}f(x)=2(-1)+1=-1.$

Continuity at $-1$ therefore requires

$-a+b=-1. \qquad (1)$

Condition at $x=2$. The middle line also covers $x=2$, so $f(2)=2a+b=\lim_{x\to 2^{-}}f(x)$, while from the right

$\lim_{x\to 2^{+}}f(x)=2^{2}-3=1.$

Continuity at $2$ therefore requires

$2a+b=1. \qquad (2)$

Solve the system. Subtract $(1)$ from $(2)$:

$\begin{aligned} (2a+b)-(-a+b) &= 1-(-1)\\ 3a &= 2\\ a &= \frac{2}{3}. \end{aligned}$

Then from $(1)$,

$b=-1+a=-1+\frac{2}{3}=-\frac{1}{3}.$

Check both equations. $-a+b=-\dfrac{2}{3}-\dfrac{1}{3}=-1$, which matches $2(-1)+1=-1$. And $2a+b=\dfrac{4}{3}-\dfrac{1}{3}=1$, which matches $2^{2}-3=1$. Both hold.

Answer. $a=\dfrac{2}{3}$ and $b=-\dfrac{1}{3}$, so the middle piece is $f(x)=\dfrac{2x-1}{3}$. With these values $f$ is continuous on all of $(-\infty,\infty)$; for any other pair, $f$ has a jump discontinuity at $x=-1$, or at $x=2$, or at both.

Identify the form. As $x\to 0$, the numerator $\sqrt{x+4}-2\to\sqrt{4}-2=0$ and the denominator $\to 0$: the indeterminate form $\dfrac{0}{0}$. So the limit may still exist, but algebra is needed to find it.

Rationalise the numerator. For $x\ne 0$ with $x>-4$, multiply above and below by $\sqrt{x+4}+2$, which is never $0$ (indeed $\sqrt{x+4}+2\ge 2$):

$\begin{aligned} \frac{\sqrt{x+4}-2}{x} &= \frac{\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)}{x\left(\sqrt{x+4}+2\right)}\\ &= \frac{(x+4)-4}{x\left(\sqrt{x+4}+2\right)}\\ &= \frac{x}{x\left(\sqrt{x+4}+2\right)}\\ &= \frac{1}{\sqrt{x+4}+2}. \end{aligned}$

Cancelling $x$ is legitimate precisely because $x\ne 0$ throughout the limit process.

Take the limit. The function $\dfrac{1}{\sqrt{x+4}+2}$ is continuous at $x=0$ — it is a quotient of continuous functions whose denominator there equals $\sqrt{4}+2=4\ne 0$ — so we may substitute:

$\lim_{x\to 0}f(x)=\frac{1}{\sqrt{0+4}+2}=\frac{1}{2+2}=\frac{1}{4}.$

Numerical check: at $x=0.001$ the original quotient is $\approx 0.24998$, and at $x=-0.001$ it is $\approx 0.25002$. Both sit next to $0.25$.

Conclusion. The limit exists and is finite while $f(0)$ is undefined, so $x=0$ is a removable discontinuity and the repairing value is

$k=\frac{1}{4}.$

The extension. Define $F(x)=f(x)$ for $x\ne 0$ and $F(0)=\dfrac{1}{4}$. The computation above shows that in fact

$F(x)=\frac{1}{\sqrt{x+4}+2} \qquad\text{for every } x>-4,$

and $\sqrt{x+4}+2$ is continuous with value at least $2$, hence never $0$, so this quotient is continuous at every $x>-4$. Therefore $F$ is continuous on $(-4,\infty)$, which is the domain of $f$ together with the single repaired point $x=0$.

(Strictly, $F$ is not continuous at $x=-4$, because $-4$ is not in its domain. The formula $\dfrac{1}{\sqrt{x+4}+2}$ does make sense there, with value $\dfrac{1}{\sqrt{0}+2}=\dfrac{1}{2}$, so one may extend once more to a function continuous on $[-4,\infty)$ — right-continuity being all that is required at that endpoint. But that is a second extension, beyond the one the question asks for.)

Set up. Let $f(x)=x^{3}-4x+1$.

Hypothesis check. $f$ is a polynomial, and every polynomial is continuous on $(-\infty,\infty)$; in particular $f$ is continuous on the closed, bounded interval $[0,1]$.

Endpoint values.

$f(0)=0-0+1=1>0, \qquad f(1)=1-4+1=-2<0.$

The target value $N=0$ lies strictly between $f(1)=-2$ and $f(0)=1$. By the Intermediate Value Theorem there exists $c$ in $(0,1)$ with $f(c)=0$, that is, $c$ solves $x^{3}-4x+1=0$.

Bisection step 1. Evaluate at the midpoint $x=\dfrac{1}{2}$:

$f\left(\frac{1}{2}\right)=\frac{1}{8}-4\cdot\frac{1}{2}+1=\frac{1}{8}-2+1=-\frac{7}{8}<0.$

Now $f(0)=1>0$ and $f\left(\dfrac{1}{2}\right)=-\dfrac{7}{8}<0$, and $f$ is continuous on $\left[0,\dfrac{1}{2}\right]$, so the IVT gives a root in $\left(0,\dfrac{1}{2}\right)$. On the other half both endpoint values are negative, so the IVT gives no information there and we discard it.

Bisection step 2. Evaluate at the midpoint $x=\dfrac{1}{4}$:

$f\left(\frac{1}{4}\right)=\frac{1}{64}-4\cdot\frac{1}{4}+1=\frac{1}{64}-1+1=\frac{1}{64}>0.$

Now $f\left(\dfrac{1}{4}\right)=\dfrac{1}{64}>0$ and $f\left(\dfrac{1}{2}\right)=-\dfrac{7}{8}<0$, with $f$ continuous on $\left[\dfrac{1}{4},\dfrac{1}{2}\right]$. By the IVT there is a root

$c \in \left(\frac{1}{4},\frac{1}{2}\right),$

an interval of length $\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$, as required.

Verification. Numerically the root is $c\approx 0.25410$, which indeed lies in $(0.25,\ 0.5)$. Note that $f$ has two further real roots, near $-2.115$ and $1.861$, both outside $[0,1]$ — the IVT never claimed uniqueness.

Rearrange. A solution of $\cos x=x$ is exactly a zero of

$g(x)=\cos x - x.$

Hypothesis check. $\cos x$ is continuous on $(-\infty,\infty)$ and $x$ is a polynomial, hence continuous; a difference of continuous functions is continuous. So $g$ is continuous on the closed, bounded interval $\left[0,\dfrac{\pi}{2}\right]$.

Endpoint values.

$g(0)=\cos 0-0=1-0=1>0,$

$g\left(\frac{\pi}{2}\right)=\cos\frac{\pi}{2}-\frac{\pi}{2}=0-\frac{\pi}{2}=-\frac{\pi}{2}\approx -1.5708<0.$

Apply the IVT. The value $N=0$ lies strictly between $g\left(\dfrac{\pi}{2}\right)=-\dfrac{\pi}{2}$ and $g(0)=1$, and $g$ is continuous on $\left[0,\dfrac{\pi}{2}\right]$. Hence there is some $c$ in $\left(0,\dfrac{\pi}{2}\right)$ with $g(c)=0$, that is,

$\cos c = c.$

Why the IVT does not give uniqueness. Its conclusion is literally "there exists at least one $c$". Nothing in the hypotheses — continuity, plus a value lying between the endpoint values — restricts how many such points there are. For example $h(x)=(x-1)(x-2)(x-3)$ is continuous on $[0,4]$ with