Corrected exercises: continuity and the Intermediate Value Theorem (MATH 140)

a) At $a = -3$ there is an open dot and no filled dot above or below it, so $f(-3)$ is undefined; on both sides the line climbs to the open dot, so $\lim_{x\to -3^-} f(x) = \lim_{x\to -3^+} f(x) = -1$. The first condition fails. At $a = -1$ the filled dot gives $f(-1) = 3$; from the left the line arrives at the open dot $(-1, 1)$, from the right the curve starts at the filled dot: $\lim_{x\to -1^-} f(x) = 1$ and $\lim_{x\to -1^+} f(x) = 3$. Condition 1 holds, condition 2 fails because the one-sided limits differ. At $a = 1$ the filled dot is at height $4$ while the curve passes through the open dot $(1, 2)$ from both sides: $f(1) = 4$ and both one-sided limits equal $2$. Conditions 1 and 2 hold and condition 3 fails, since $2 \ne 4$. At $a = 3$, $f(3)$ is undefined, the curve plunges to $-\infty$ on the left and comes down from $+\infty$ on the right. Condition 1 fails. Read the value from the DOTS and the limits from the CURVE: the filled dot at $(1, 4)$ says nothing about where the curve is heading.

b) At $-3$ the limit exists (it is $-1$) and only the value is missing: removable. At $1$ the limit exists (it is $2$) but the value is wrong: removable as well. At $-1$ the one-sided limits are finite and different: jump. At $3$ both one-sided limits are infinite: infinite discontinuity. The trap is to classify by the condition that fails first: at $-3$ and at $3$ it is the same condition, the value is undefined, and yet one discontinuity is removable and the other is infinite. The TYPE is decided by the one-sided limits, never by the list of failed conditions.

c) Continuity from the left at $-1$ asks $\lim_{x\to -1^-} f(x) = f(-1)$, that is $1 = 3$: false. Continuity from the right asks $\lim_{x\to -1^+} f(x) = f(-1)$, that is $3 = 3$: true. So $f$ is continuous from the right at $-1$ and not from the left; the filled dot sits on the right-hand branch. At $x = -4$ the function is not defined to the left, so the only question that makes sense is continuity from the right: $\lim_{x\to -4^+} f(x) = -2 = f(-4)$, and $f$ is continuous from the right at $-4$. This is exactly what continuity on a closed interval asks at its endpoints.

d) A removable discontinuity is repaired by one value, the limit. At $-3$, defining $f(-3) = -1$ makes all three conditions hold. At $1$, redefining $f(1) = 2$ instead of $4$ does the same. At $-1$ no value can work: whatever $f(-1)$ is, it cannot equal both $1$ and $3$, so condition 2 still fails; changing one value never closes a gap between two different one-sided limits. At $3$ there is no finite limit, so condition 2 fails whatever value is chosen. The question is always answered by the limit, and only a limit that exists and is finite can be used.

e) $f$ is continuous on $[-4, -3)$, $(-3, -1)$, $[-1, 1)$, $(1, 3)$ and $(3, 5]$: each discontinuity is excluded, and an endpoint is kept exactly when $f$ has the one-sided continuity it needs there (from the right at $-4$ and at $-1$, from the left at $5$, where $f(5) = 2$ is the end of the curve). On $[-1, 1)$, $f$ is continuous at every interior point and from the right at $-1$: yes. On $[-1, 1]$, continuity at the right endpoint $1$ would require $\lim_{x\to 1^-} f(x) = f(1)$, that is $2 = 4$: no. A closed interval makes a demand at each end, and those demands are one-sided.

a) Factor first: $f(x) = \frac{(x - 2)(x + 2)}{(x - 1)(x - 2)}$, with domain $\mathbb{R}$ minus $\{1, 2\}$. As a rational function, $f$ is continuous at every other point, so only $1$ and $2$ are candidates. At $2$: for $x \ne 2$, $f(x) = \frac{x + 2}{x - 1}$, which is continuous at $2$, so $\lim_{x\to 2} f(x) = \frac{4}{1} = 4$. The limit exists and $f(2)$ is undefined: removable discontinuity. At $1$: in the simplified form, the numerator $x + 2$ tends to $3$ while $x - 1 \to 0$; for $x \to 1^+$, $\frac{x+2}{x-1}$ is a number close to $3$ divided by a small positive number, so $\lim_{x\to 1^+} f(x) = \infty$. One infinite one-sided limit is enough: infinite discontinuity.

b) The domain excludes $1$ only, and $g$ is continuous elsewhere. An absolute value is removed by cases. For $x > 1$, $|x - 1| = x - 1$ and $g(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \to 2$. For $x < 1$, $|x - 1| = -(x - 1)$ and $g(x) = -(x + 1) \to -2$. The one-sided limits are finite and different: jump discontinuity at $1$, and no value $g(1)$ can repair it. Writing $\frac{x^2 - 1}{|x - 1|} = x + 1$ without the cases gives the limit $2$ and a false removable discontinuity: the absolute value changes the sign on one side only, and that is the whole point.

c) Between two integers, $\lfloor x \rfloor$ is constant, so $h(x) = x - n$ on $(n, n + 1)$ is a polynomial there and $h$ is continuous at every non-integer. At an integer $n$: for $n \le x < n + 1$, $h(x) = x - n \to 0$ as $x \to n^+$; for $n - 1 \le x < n$, $\lfloor x \rfloor = n - 1$ and $h(x) = x - n + 1 \to 1$ as $x \to n^-$. And $h(n) = n - n = 0$. The one-sided limits $1$ and $0$ differ: a jump at every integer. Since $\lim_{x\to n^+} h(x) = 0 = h(n)$, $h$ is continuous from the right at $n$ and not from the left. The trap is to compute $\lfloor x \rfloor$ to the left of $n$ as $n$: just below $3$, $\lfloor 2.99 \rfloor = 2$.

d) For $k$: $\sin\frac{1}{x} = 1$ when $\frac{1}{x} = \frac{\pi}{2} + 2j\pi$, that is at $x = \frac{2}{(4j + 1)\pi}$, and $\sin\frac{1}{x} = -1$ at $x = \frac{2}{(4j + 3)\pi}$, for $j = 1, 2, 3, \dots$: both families of points come as close to $0$ as we like, on the right (and, by oddness, on the left). So $k$ has no limit from either side, and since $|k| \le 1$ neither one-sided limit is infinite. This discontinuity is none of the three types: the classification of Stewart is not a complete list. No value $k(0)$ helps, because condition 2 fails. For $m$: $-|x| \le x\sin\frac{1}{x} \le |x|$ for $x \ne 0$, since $|\sin| \le 1$, and both bounds tend to $0$, so by the squeeze theorem $\lim_{x\to 0} m(x) = 0$. The limit exists and $m(0)$ is undefined: removable, and $m(0) = 0$ makes $m$ continuous, as the figure of the solution shows.

e) A vertical asymptote needs an infinite one-sided limit, and a zero of the denominator does not guarantee one. At $2$ the numerator vanishes too, the factor $x - 2$ cancels, and the limit is the finite number $4$: the graph has a HOLE at $(2, 4)$, not an asymptote. At $1$ only the denominator vanishes, and the asymptote is real. Correct statement: $f$ is discontinuous at $1$ and at $2$, with an infinite discontinuity (a vertical asymptote) at $1$ and a removable one at $2$. Simplify the common factors BEFORE naming any asymptote.

a) The root requires $4 - x^2 \ge 0$, that is $-2 \le x \le 2$, and the quotient requires $x \ne 1$. On this domain, $4 - x^2$ is a polynomial, so continuous, and $\sqrt{\ }$ is continuous on $[0, \infty)$, so $\sqrt{4 - x^2}$ is continuous on $[-2, 2]$ by composition; $x - 1$ is continuous and non-zero for $x \ne 1$, so the quotient is continuous on $[-2, 1)$ and $(1, 2]$. At $-2$ and $2$ the function is defined on one side only, and the continuity there is one-sided: from the right at $-2$, from the left at $2$, with $F(\pm 2) = 0$. At $1$, $F$ is undefined. Answering $\mathbb{R}$ minus $\{1\}$ forgets the root; answering $(-2, 1) \cup (1, 2)$ throws away two endpoints where $F$ is perfectly continuous.

b) $G$: the logarithm is continuous on $(0, \infty)$, so $G$ is continuous wherever $x^2 - 5x + 6 = (x - 2)(x - 3) > 0$, that is for $x < 2$ or $x > 3$ (the product of two factors is positive outside the roots). $G$ is continuous on $(-\infty, 2)$ and $(3, \infty)$, and undefined on $[2, 3]$. $H$: $\sqrt x$ needs $x \ge 0$, $e^u$ is continuous everywhere, so the numerator is continuous on $[0, \infty)$; the denominator is continuous and vanishes exactly when $\cos x = -1$, that is at $x = \pi + 2j\pi$. So $H$ is continuous on $[0, \pi)$, $(\pi, 3\pi)$, $(3\pi, 5\pi)$, and so on: every odd multiple of $\pi$ is excluded, and $0$ is kept with continuity from the right.

c) Let $g(x) = x + \sin x$: a sum of continuous functions, so $\lim_{x\to\pi} g(x) = g(\pi) = \pi + 0 = \pi$. The sine is continuous at $\pi$, so $\lim_{x\to\pi} \sin(g(x)) = \sin \pi = 0$. For the second limit, the inner function $q(x) = \frac{x^2 - 4}{4x - 8}$ is NOT defined at $2$, so direct substitution is impossible, but its limit exists: $q(x) = \frac{(x - 2)(x + 2)}{4(x - 2)} = \frac{x + 2}{4} \to 1$. The arctangent is continuous at $1$, so the limit is $\arctan 1 = \frac{\pi}{4}$. This is the limit form of the composition theorem: the inner function needs only a limit, the OUTER function needs continuity at that limit.

d) For $|x| < 1$ we have $0 \le x^2 < 1$, so $\lfloor x^2 \rfloor = 0$, which is also its value at $0$: $\lfloor x^2 \rfloor$ is constant near $0$, hence continuous there, although $\lfloor\ \rfloor$ is discontinuous at $0$. For $0 < |x| < 1$, $-1 < -x^2 < 0$, so $\lfloor -x^2 \rfloor = -1$, while $\lfloor -0 \rfloor = 0$: the limit is $-1$ and the value is $0$, a removable discontinuity. The composition theorem gives a SUFFICIENT condition only. When the outer function is discontinuous at $b$, anything can happen, and here the answer depends on the side from which the inner function reaches $0$: $x^2$ arrives from above, where $\lfloor\ \rfloor$ is continuous from the right, and $-x^2$ from below, where it jumps.

a) For $x < 1$, $f$ is a polynomial, and for $x > 1$ as well, so $f$ is continuous at every $x \ne 1$. At the seam: $\lim_{x\to 1^-} f(x) = k + 3$ (the left polynomial is continuous at $1$), and $\lim_{x\to 1^+} f(x) = f(1) = 2 - k$. Continuity at $1$ requires $k + 3 = 2 - k$, so $2k = -1$ and $k = -\frac{1}{2}$. Check: left $-\frac{1}{2} + 3 = \frac{5}{2}$, right $2 + \frac{1}{2} = \frac{5}{2}$. The value $f(1)$ is given by the piece that contains the equality sign, here $x \ge 1$, and it must match the OTHER side's limit.

b) The left piece is continuous on $(-\infty, 1)$ (its denominator vanishes only at $1$, which is excluded), the middle one is a polynomial, and the right piece is continuous on $(3, \infty)$. Seam at $1$: $g(1) = a + b$ and $\lim_{x\to 1^+} g(x) = a + b$; on the left, plugging in gives $\frac{0}{0}$, so simplify: $\frac{(x - 1)(x + 1)}{x - 1} = x + 1 \to 2$. Condition: $a + b = 2$. Seam at $3$: $g(3) = 3a + b$ and $\frac{x^2 - x - 6}{x - 3} = \frac{(x - 3)(x + 2)}{x - 3} = x + 2 \to 5$ as $x \to 3^+$. Condition: $3a + b = 5$. Subtracting, $2a = 3$, so $a = \frac{3}{2}$ and $b = \frac{1}{2}$. Check: $g(1) = 2$, $g(3) = \frac{9}{2} + \frac{1}{2} = 5$; the figure shows the three pieces forming one unbroken line. A student who plugs $x = 1$ into the left piece and writes that $g$ cannot be continuous because of the $\frac{0}{0}$ has confused the value of a formula at a point with the limit of the function there.

c) Both pieces are polynomials, so only $x = 1$ matters: $\lim_{x\to 1^-} h(x) = c^2$ and $\lim_{x\to 1^+} h(x) = h(1) = 3c - 2$. Condition: $c^2 = 3c - 2$, that is $c^2 - 3c + 2 = (c - 1)(c - 2) = 0$, so $c = 1$ or $c = 2$. Both work: with $c = 1$ the two sides give $1$ and $1$; with $c = 2$ they give $4$ and $4$. Dividing the equation by $c$, or keeping only the first root found, loses a mark: when the parameter enters squared, the answer is a SET of values, and each one is checked.

d) For $x > 0$, $|x| = x$ and $p(x) = \frac{x(x + 3)}{x} = x + 3 \to 3$. For $x < 0$, $|x| = -x$ and $p(x) = -(x + 3) \to -3$. The one-sided limits are finite and different, so $\lim_{x\to 0} p(x)$ does not exist and condition 2 fails whatever $c$ is: there is NO value of $c$. This is a jump discontinuity, and the value at one point cannot close a gap between two sides. The expected answer is a sentence, not a number: a student who solves $c = 3$ from the right side alone has made $p$ continuous from the right only.

a) Let $f(x) = x^5 - 3x + 1$, a polynomial, hence continuous on $\mathbb{R}$ and on every closed interval. Exact values: $f(-2) = -32 + 6 + 1 = -25$, $f(-1) = -1 + 3 + 1 = 3$, $f(0) = 1$, $f(1) = 1 - 3 + 1 = -1$, $f(2) = 32 - 6 + 1 = 27$. The sign changes on $[-2, -1]$, on $[0, 1]$ and on $[1, 2]$, so by the IVT with $N = 0$ there is a root in each of $(-2, -1)$, $(0, 1)$ and $(1, 2)$. These open intervals are disjoint, so the three roots are distinct: at least three real solutions. Note the words AT LEAST: the IVT counts sign changes, not roots, and a degree-5 polynomial could have up to five real roots. The figure of the solution shows the three crossings; $f$ has no fourth one, but proving it is not an IVT question.