Corrected exercises: continuity (MATH 203)

The filled dot above $x = -2$ gives $f(-2) = 2$. From the left, the line climbs to the open dot $(-2, 1)$: $\lim_{x\to -2^-} f(x) = 1$. From the right, the V-shaped piece starts at the open dot $(-2, 4)$: $\lim_{x\to -2^+} f(x) = 4$. Condition 1 holds, since the value exists; condition 2 fails, since the one-sided limits differ. Read the VALUE from the dots and the LIMITS from where the curve is heading: the filled dot sits between the two ends, on neither branch, and it says nothing about either limit.

At $1$ there is only an open dot, at the bottom of the V, and no filled dot above or below it: $f(1)$ is undefined and condition 1 fails. The curve reaches the height $1$ from both sides, so $\lim_{x\to 1} f(x) = 1$ exists. At $3$ the filled end of the V is $(3, 3)$, so $f(3) = 3$; from the right, the next piece starts at the same height, so both one-sided limits equal $3$. The three conditions hold: $f$ IS continuous at $3$. A change of formula, or a sharp turn of the graph, is not a break: continuity only asks that the pieces meet at the value of the function.

Approaching $4$ from the left, the curve climbs along the asymptote without bound: $\lim_{x\to 4^-} f(x) = +\infty$. From the right, the last segment starts at the filled dot $(4, 1)$: $\lim_{x\to 4^+} f(x) = 1 = f(4)$. Continuity from the left would require $\lim_{x\to 4^-} f(x) = f(4)$, impossible with an infinite limit: no. Continuity from the right requires $\lim_{x\to 4^+} f(x) = f(4)$, that is $1 = 1$: yes. An asymptote on ONE side does not prevent a value, nor continuity from the other side.

At $-2$ the one-sided limits are finite and different: a jump discontinuity. The isolated value $f(-2) = 2$ does not change the type, and no value can repair it, since no number equals both $1$ and $4$. At $1$ the limit exists and only the value is missing: removable, and defining $f(1) = 1$ makes the three conditions hold. At $4$ a one-sided limit is infinite: infinite discontinuity, and no finite value repairs it. Only the removable discontinuity can be repaired, and only by the value of the limit.

Each discontinuity cuts the domain. At $-4$, $f$ is continuous from the right ($f(-4) = -1$ is where the line starts), so $-4$ is kept. At $-2$, $f$ is continuous from neither side ($1 \ne 2$ and $4 \ne 2$), so $-2$ is excluded from both intervals around it. At $1$ the function is undefined. At $4$, $f$ is continuous from the right, so $4$ opens the last interval, which ends at $6$ with continuity from the left. Answer: $[-4, -2)$, $(-2, 1)$, $(1, 4)$ and $[4, 6]$. The interval $(1, 4)$ contains the seam at $3$, where b) showed continuity.

Group the denominator: $x^3 + 3x^2 - 4x - 12 = x^2(x + 3) - 4(x + 3) = (x + 3)(x^2 - 4) = (x + 3)(x - 2)(x + 2)$. The numerator is $(x + 3)(x - 2)$. The candidates are $-3$, $2$ and $-2$, and for $x \ne -3, 2$, $f(x) = \frac{1}{x + 2}$. At $-3$: $\lim_{x\to -3} f(x) = \frac{1}{-1} = -1$, finite, $f(-3)$ undefined: removable. At $2$: the limit is $\frac{1}{4}$: removable. At $-2$: the numerator tends to $(1)(-4) = -4$ while the denominator tends to $0$, so $|f(x)| \to \infty$: infinite discontinuity. Stopping after $x^2(x + 3) - 4(x + 3)$, which is still a SUM, finds no common factor at all and misses both removable points.

Factor the trinomial with its leading coefficient: $2x^2 - 5x + 3 = (2x - 3)(x - 1)$, checked by expanding, $2x^2 - 2x - 3x + 3$. The denominator is a difference of cubes: $x^3 - 1 = (x - 1)(x^2 + x + 1)$, and $x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so it never vanishes. The only candidate is $1$, and for $x \ne 1$, $g(x) = \frac{2x - 3}{x^2 + x + 1}$, continuous at $1$: $\lim_{x\to 1} g(x) = \frac{-1}{3} = -\frac{1}{3}$. Removable discontinuity. Two classic slips: $x^3 - 1 = (x - 1)^3$ (expand it: $x^3 - 3x^2 + 3x - 1$), and a factorization $(2x - 1)(x - 3)$ that gives $-7x$ in the middle instead of $-5x$.

$h$ is undefined at $0$ (because of $\frac{1}{x}$) and at $3$. Clear the complex fraction by multiplying top and bottom by $3x$: $h(x) = \frac{3 - x}{3x(x - 3)}$. Now $3 - x = -(x - 3)$, so for $x \ne 0, 3$, $h(x) = -\frac{1}{3x}$. At $3$: $\lim_{x\to 3} h(x) = -\frac{1}{9}$, removable. At $0$: $-\frac{1}{3x}$ is unbounded, infinite discontinuity. The sign is the whole point: writing $\frac{3 - x}{x - 3} = 1$ gives $+\frac{1}{9}$, a wrong answer that looks perfectly reasonable. Check with the calculator: $h(3.001) = \frac{1/3.001 - 1/3}{0.001} \approx -0.1111$.

Factor the denominator: $x^2 - 2x - 3 = (x - 3)(x + 1)$, so the candidates are $3$ and $-1$. Near $3$, $x - 1 > 0$, so $|x - 1| = x - 1$, the numerator is $x - 3$, and $k(x) = \frac{x - 3}{(x - 3)(x + 1)} = \frac{1}{x + 1} \to \frac{1}{4}$: removable. Near $-1$, $x - 1 < 0$, so $|x - 1| = 1 - x$, the numerator is $1 - x - 2 = -(x + 1)$, and $k(x) = \frac{-(x + 1)}{(x - 3)(x + 1)} = -\frac{1}{x - 3} \to -\frac{1}{-4} = \frac{1}{4}$: removable as well. Writing $|x - 1| = x - 1$ everywhere gives $\frac{1}{x + 1}$ near $-1$ and a false infinite discontinuity. At $1$, $|x - 1|$ and $x^2 - 2x - 3$ are continuous and the denominator equals $-4 \ne 0$, so $k$ is continuous by the quotient rule, with $k(1) = \frac{-2}{-4} = \frac{1}{2}$: the corner of $|x - 1|$ is not a discontinuity.

The cancellation shows that $f$ and $\frac{1}{x + 2}$ agree for $x \ne -3, 2$, which is exactly what a limit needs: $\lim_{x\to 2} f(x) = \frac{1}{4}$. It does not make $f(2)$ exist: the original formula gives $\frac{0}{0}$, so condition 1 fails and $f$ is discontinuous at $2$. Correct statement: $f$ has a removable discontinuity at $2$, and its continuous extension, with the value $\frac{1}{4}$ at $2$ (and $-1$ at $-3$), is the function $\frac{1}{x + 2}$ on its own domain. The domain is decided on the ORIGINAL formula, before any cancellation.

The root needs $6 - 2x \ge 0$. Solve it with care: $-2x \ge -6$, and dividing by $-2$ REVERSES the inequality, $x \le 3$. The denominator factors as $x^2 - x - 2 = (x - 2)(x + 1)$, so $x \ne 2$ and $x \ne -1$. On this domain, $6 - 2x$ is a polynomial and the square root is continuous on $[0, \infty)$, so the numerator is continuous on $(-\infty, 3]$ by composition; the denominator is a polynomial, non-zero away from $-1$ and $2$; the quotient is continuous on $(-\infty, -1)$, $(-1, 2)$ and $(2, 3]$. At $3$, $F$ is defined on the left only and $F(3) = 0$: the continuity there is from the left. At $-1$ and $2$, $F$ is undefined. Writing $x \ge 3$ after forgetting the reversal puts the whole answer on the wrong side of $3$.

The cosine and the sine are continuous everywhere, so $H$ is continuous wherever $2\sin x - 1 \ne 0$, that is $\sin x \ne \frac{1}{2}$. On $[0, 2\pi]$ the unit circle gives $\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ (first quadrant) and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (second quadrant), and nowhere else. So $H$ is continuous on $\left[0, \frac{\pi}{6}\right)$, $\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)$ and $\left(\frac{5\pi}{6}, 2\pi\right]$, one-sided at $0$ and $2\pi$. The second solution is the one that goes missing: $\sin$ is positive in two quadrants.

Inner function first: for $x \ne 0$, $\frac{\pi x^2 + \pi x}{3x} = \frac{\pi x(x + 1)}{3x} = \frac{\pi(x + 1)}{3}$, so its limit at $0$ is $\frac{\pi}{3}$, although it is undefined at $0$. The cosine is continuous at $\frac{\pi}{3}$, so by the limit form of the composition theorem the limit is $\cos\frac{\pi}{3} = \frac{1}{2}$. Second limit: $\frac{x^3 - 27}{x - 3} = x^2 + 3x + 9$ for $x \ne 3$ (difference of cubes), which tends to $9 + 9 + 9 = 27$. The logarithm is continuous at $27 > 0$, so the limit is $\ln 27 = 3\ln 3 \approx 3.2958$. The inner function only needs a LIMIT; the outer function needs CONTINUITY at that limit.

$f \circ g(x) = f(x^2) = \frac{1}{x^2 - 1}$. The function $f$ fails at $u = 1$, so $f \circ g$ fails where $g(x) = 1$, that is $x^2 = 1$: at $x = -1$ AND $x = 1$ (infinite discontinuities). The trap is to copy the bad point of $f$, $1$, as the bad point of $f \circ g$: the question is which $x$ are SENT to the bad point, and that is an equation to solve. In the other order, $g \circ f(u) = \left(\frac{1}{u - 1}\right)^2 = \frac{1}{(u - 1)^2}$ is discontinuous at $u = 1$ only, where $f$ itself is undefined; $g$ is continuous everywhere and adds no bad point. Composition does not commute, and neither do its discontinuities.

For $\frac{1}{2} \le x < 5$ the first piece is continuous (a root of a positive-or-zero polynomial over a non-zero polynomial), and the second is a polynomial. At the seam $5$: $f(5) = 25k - 8 = \lim_{x\to 5^+} f(x)$. On the left, the form is $\frac{0}{0}$; multiply by the conjugate $\sqrt{2x - 1} + 3$: the numerator becomes $(2x - 1) - 9 = 2x - 10 = 2(x - 5)$, so $f(x) = \frac{2}{\sqrt{2x - 1} + 3} \to \frac{2}{3 + 3} = \frac{1}{3}$. Condition: $25k - 8 = \frac{1}{3}$, so $25k = \frac{25}{3}$ and $k = \frac{1}{3}$. The factor $2$ in $2x - 10$ is the mark: forgetting it gives $\frac{1}{6}$ and $k = \frac{49}{150}$, a number that should alarm you.

The pieces are continuous on $(-\infty, -1)$ (the only zeros of denominators, $0$ and $-1$, are not in it), on $(-1, 2)$ and on $(2, \infty)$ (there $\sqrt{x + 2} - 2 > 0$). Seam at $-1$: $g(-1) = -a + b$. The left piece is a complex fraction: multiply top and bottom by $x$ to get $\frac{2 + 2x}{x(x + 1)} = \frac{2(x + 1)}{x(x + 1)} = \frac{2}{x}$, which tends to $-2$. Condition: $-a + b = -2$. Seam at $2$: $g(2) = 2a + b$, and on the right multiply by the conjugate: $\frac{(x - 2)(\sqrt{x + 2} + 2)}{(x + 2) - 4} = \sqrt{x + 2} + 2 \to 4$. Condition: $2a + b = 4$. Subtracting, $3a = 6$, so $a = 2$ and $b = 0$. Check: $2(-1) = -2$ and $2(2) = 4$; the figure shows three pieces forming one unbroken graph.

Both pieces are polynomials, so only $3$ matters: $\lim_{x\to 3^-} h(x) = 9 - c^2$ and $h(3) = \lim_{x\to 3^+} h(x) = 3c - 1$. Condition: $9 - c^2 = 3c - 1$. Move everything to ONE side, keeping the signs: $c^2 + 3c - 10 = 0$, that is $(c + 5)(c - 2) = 0$. So $c = 2$ or $c = -5$. Check both: with $c = 2$, $9 - 4 = 5$ and $6 - 1 = 5$; with $c = -5$, $9 - 25 = -16$ and $-15 - 1 = -16$. The negative value looks strange and is perfectly correct; a student who writes $c^2 - 3c - 10 = 0$ (a sign lost in the transfer) finds $5$ and $-2$, and the check exposes it at once: $9 - 25 \ne 15 - 1$.

Factor both: $x^2 - 1 = (x - 1)(x + 1)$ and $x^2 - 2x + 1 = (x - 1)^2$. One factor $x - 1$ cancels and ONE REMAINS: $q(x) = \frac{x + 1}{x - 1}$ for $x \ne 1$. As $x \to 1$ the numerator tends to $2$ and the denominator to $0$, so $q(x) \to -\infty$ from the left and $+\infty$ from the right. The limit does not exist as a finite number, condition 2 fails whatever $c$ is: there is NO value of $c$. Seeing a common factor and concluding removable, then computing $c$ from the leftover formula, is the trap: count the multiplicity of the factor before cancelling.