Corrected exercises: one-sided limits (MATH 203)

a) To the left of $-1$ the graph is the rising parabola piece, which climbs to the EMPTY dot at height $2$: $\lim_{x\to -1^-} g(x) = 2$. To the right of $-1$ the graph is the straight piece that starts at the FULL dot $(-1, 0)$: $\lim_{x\to -1^+} g(x) = 0$. The one-sided limits exist and differ, so $\lim_{x\to -1} g(x)$ does not exist. The full dot gives $g(-1) = 0$, which equals the right-hand limit; that does not rescue the two-sided limit, which needs the LEFT side to agree too.

b) From the left, the straight piece runs up to the empty dot at height $1$; from the right, the second parabola piece leaves from that same empty dot. So $\lim_{x\to 1^-} g(x) = 1 = \lim_{x\to 1^+} g(x)$, and therefore $\lim_{x\to 1} g(x) = 1$. The isolated full dot at height $3$ gives the value $g(1) = 3$, a different question. Answering $3$ for the limit because that dot is full is the most common reading error of the chapter: a limit reads the curve AROUND $1$, never the dot AT $1$.

c) From the left the parabola piece ends at the empty dot $(3, 3)$, so $\lim_{x\to 3^-} g(x) = 3$; from the right the line $y = 4 - x$ leaves from the empty dot $(3, 1)$, so $\lim_{x\to 3^+} g(x) = 1$. The sides differ: $\lim_{x\to 3} g(x)$ does not exist. Both dots are empty, so $g(3)$ is not defined. Two separate facts: a limit may fail where the function is defined ($a = -1$), and exist where it is not defined (a hole); here both happen to fail.

d) At $-3$ the graph only lives to the RIGHT: starting at the full dot $(-3, 0)$, $\lim_{x\to -3^+} g(x) = 0$. At $5$ it only lives to the LEFT: the line comes down to the full dot $(5, -1)$, so $\lim_{x\to 5^-} g(x) = -1$. Since $g$ is not defined to the right of $5$, there is no right-hand limit there, and in Thomas's convention the two-sided limit $\lim_{x\to 5} g(x)$ does not exist; the one-sided limit $\lim_{x\to 5^-} g(x) = -1$ is the complete answer at that endpoint. Writing $\lim_{x\to 5^+} g(x) = -1$ is a limit on a side where $g$ has no values.

e) Inside $(-3, 5)$ the graph is unbroken except at $-1$, $1$ and $3$. At $a = -1$ and $a = 3$ the one-sided limits differ, so the limit does not exist. At $a = 1$ the limit exists, equal to $1$, but differs from $g(1) = 3$. Everywhere else the limit exists and equals $g(a)$.

a) The inside $x - 3$ vanishes at $3$: split. Factor first, $x^2 - 9 = (x - 3)(x + 3)$. For $x > 3$, $|x - 3| = x - 3$ and the quotient is $x + 3 \to 6$. For $x < 3$, $x - 3 < 0$, so $|x - 3| = -(x - 3)$ and the quotient is $-(x + 3) \to -6$. The sides differ, so $\lim_{x\to 3} \frac{x^2 - 9}{|x - 3|}$ does not exist. The lost marks are on the left: writing $|x - 3| = 3 - x$ is right, writing $|x - 3| = x - 3$ “because absolute values remove the minus” is not.

b) For $x > 1$: $|x - 1| = x - 1$, the numerator is $2(x - 1)$ and the quotient is $2$. For $x < 1$: $|x - 1| = -(x - 1)$, the numerator is $-(x - 1) + (x - 1) = 0$, and the quotient is $0$ for every $x < 1$. So $\lim_{x\to 1^-} = 0$ and $\lim_{x\to 1^+} = 2$: no two-sided limit. On the left the numerator is identically $0$; a student who does not simplify first sees $\frac{0}{0}$ and panics.

c) The inside $x$ vanishes at $0$, so check both sides, but first use $x^2 = |x|^2$: for $x \ne 0$, $\frac{|x|}{|x|^2 + |x|} = \frac{1}{|x| + 1}$, which tends to $1$ from either side. The limit is $1$. Splitting also works: $\frac{x}{x^2 + x} = \frac{1}{x + 1} \to 1$ for $x > 0$ and $\frac{-x}{x^2 - x} = \frac{-1}{x - 1} \to 1$ for $x < 0$. An absolute value is a reason to CHECK both sides, never a proof that they differ.

d) Factor the numerator: $2x^2 - 5x = x(2x - 5)$. The inside $2x - 5$ vanishes at $\frac{5}{2}$. For $x > \frac{5}{2}$ the quotient is $x \to \frac{5}{2}$; for $x < \frac{5}{2}$ it is $-x \to -\frac{5}{2}$. The limit does not exist. The common factor $x$ is where the algebra goes wrong: $2x^2 - 5x$ is $x(2x - 5)$, not $2x(x - 5)$, and only the correct factorization shows the factor $2x - 5$ that cancels with the absolute value.

e) The inside of $|x|$ is $x$, which vanishes at $0$, not at $-3$. Near $-3$, $x < 0$ on BOTH sides, so $|x| = -x$ throughout and the quotient is $\frac{-x - 3}{x + 3} = -1$ for $x \ne -3$ near $-3$. Both one-sided limits equal $-1$, and so does the limit. Splitting at $-3$ with $|x| = x$ on the right would give $\frac{x - 3}{x + 3} \to \frac{-6}{0}$, nonsense: the side decides nothing, the sign of the inside decides everything.

a) For $1 \le x < 2$, $\lfloor x \rfloor = 1$ and $x - \lfloor x \rfloor = x - 1 \to 1$: on the figure, the tooth climbs to the empty dot at height $1$. For $2 \le x < 3$, $\lfloor x \rfloor = 2$ and $x - 2 \to 0$: the next tooth starts at the full dot $(2, 0)$. So the left limit is $1$, the right limit is $0$, the value is $2 - 2 = 0$, and the two-sided limit does not exist. The same happens at every integer.

b) Left of $3$, say $2.9 < x < 3$: $\lfloor x \rfloor = 2$, and $x^2$ is just BELOW $9$, above $8.41$, so $\lfloor x^2 \rfloor = 8$. The expression is $4 - 8 = -4$ on that whole interval, so the left limit is $-4$. Right of $3$, $3 < x < 3.1$: $\lfloor x \rfloor = 3$ and $9 < x^2 < 9.61$, so $\lfloor x^2 \rfloor = 9$ and the expression is $0$. The trap is to write $\lfloor x^2 \rfloor = \lfloor x \rfloor^2 = 4$ on the left: the floor is read on its own argument $x^2$, which lies in $[8, 9)$, not on $x$.

c) For $0 < x \le 1$, $\lceil x \rceil = 1$, so $\lceil x \rceil - x = 1 - x \to 0$ as $x \to 1^-$. For $1 < x \le 2$, $\lceil x \rceil = 2$, so the expression is $2 - x \to 1$ as $x \to 1^+$. The ceiling jumps just AFTER each integer, the floor AT each integer; in both cases the answer comes from the interval, not from a memorized picture.

d) The argument is $2x$. As $x \to -\frac{1}{2}^-$, $2x \to -1^-$, so $2x$ lies in $[-2, -1)$ and $\lfloor 2x \rfloor = -2$. As $x \to -\frac{1}{2}^+$, $2x$ lies in $[-1, 0)$ and $\lfloor 2x \rfloor = -1$. For $2\lfloor x \rfloor$ the argument is $x$, which stays in $(-1, 0)$ on both sides of $-\frac{1}{2}$: $\lfloor x \rfloor = -1$ and $2\lfloor x \rfloor = -2$, so the limit exists and equals $-2$. The two functions differ: at $x = -\frac{1}{4}$, $\lfloor 2x \rfloor = \lfloor -\frac{1}{2} \rfloor = -1$ while $2\lfloor x \rfloor = -2$. A constant does not move out of a floor.

e) For $1 \le x < 2$ the product is $1 \cdot (x - 1) \to 1$. For $2 \le x < 3$ it is $2(x - 2) \to 0$. Left $1$, right $0$: the limit at $2$ does not exist. Each factor is replaced by its formula ON THAT SIDE before any limit is taken; a student who uses $\lfloor 2 \rfloor = 2$ on the left gets $2 \cdot 1 = 2$.

a) The angle is $\frac{3x}{4}$, so the denominator must become $\frac{3x}{4}$. Write $5x = \frac{20}{3} \cdot \frac{3x}{4}$: $\frac{\sin(3x/4)}{5x} = \frac{3}{20} \cdot \frac{\sin(3x/4)}{3x/4}$. As $x \to 0$, $\theta = \frac{3x}{4} \to 0$, so the quotient tends to $1$ and the limit is $\frac{3}{20}$ by the constant multiple rule. The algebra that costs marks is the compound fraction: dividing by $\frac{3x}{4}$ is multiplying by $\frac{4}{3x}$, and $\frac{1}{5} \cdot \frac{3}{4} = \frac{3}{20}$, not $\frac{15}{4}$.

b) This is the RECIPROCAL of the theorem. Write $\frac{6t}{\sin 2t} = 3 \cdot \frac{2t}{\sin 2t} = \frac{3}{\frac{\sin 2t}{2t}}$. The denominator tends to $1 \ne 0$, so the quotient rule applies and the limit is $\frac{3}{1} = 3$. The theorem is about $\frac{\sin \theta}{\theta}$; its reciprocal also tends to $1$, but only because the quotient rule has its permit, a sentence worth writing.

c) First the identity $\tan 3x = \frac{\sin 3x}{\cos 3x}$. Then match each sine with its own angle: $\frac{\tan 3x}{\sin 8x} = \frac{3}{8} \cdot \frac{\sin 3x}{3x} \cdot \frac{8x}{\sin 8x} \cdot \frac{1}{\cos 3x}$. The three factors tend to $1$, $1$ and $\frac{1}{\cos 0} = 1$, so the limit is $\frac{3}{8}$ by the product rule. Writing $\tan 3x = 3\tan x$ is the classic algebra slip: a constant does not come out of a trigonometric function.

d) Split the $x^2$ between the two sines: $\frac{x^2}{\sin x \sin 4x} = \frac{x}{\sin x} \cdot \frac{x}{\sin 4x} = \frac{x}{\sin x} \cdot \frac{1}{4} \cdot \frac{4x}{\sin 4x}$. Each reciprocal quotient tends to $1$, so the limit is $\frac{1}{4}$. The missing $4$ is the whole question: $\frac{x}{\sin 4x}$ does not tend to $1$, since its angle is $4x$ and its numerator only $x$.

e) Substitution gives $\frac{0}{0}$. Put the cosine quotient in the form of the theorem: $\frac{\cos 5h - 1}{\sin 2h} = \frac{\cos 5h - 1}{5h} \cdot \frac{5h}{2h} \cdot \frac{2h}{\sin 2h}$. As $h \to 0$: the first factor tends to $0$ (angle $5h \to 0$), the second is the constant $\frac{5}{2}$, the third tends to $1$. The limit is $0 \cdot \frac{5}{2} \cdot 1 = 0$. Answering $\frac{5}{2}$ treats $\frac{\cos 5h - 1}{5h}$ as if it tended to $1$: the cosine limit is $0$, not $1$.

a) The identity turns the cosine into a square of sines: $\frac{\cos h - 1}{h} = \frac{-2\sin^2(h/2)}{h}$. Write $h = 2 \cdot \frac{h}{2}$ to match one sine with its angle: $\frac{-2\sin^2(h/2)}{2 \cdot (h/2)} = -\sin\frac{h}{2} \cdot \frac{\sin(h/2)}{h/2}$. As $h \to 0$, $\theta = \frac{h}{2} \to 0$, so the second factor tends to $1$, and the first tends to $-\sin 0 = 0$. By the product rule, the limit is $0 \cdot 1 = 0$. This is Thomas's own proof; the step that goes wrong on paper is $\frac{2}{h} = \frac{1}{h/2}$, a compound fraction.

b) Half angle with $u = 3x$: $1 - \cos 6x = 2\sin^2 3x$. So for $x$ near $0$, $x \ne 0$, $\frac{1 - \cos 6x}{x \sin 3x} = \frac{2\sin^2 3x}{x \sin 3x} = \frac{2\sin 3x}{x} = 6 \cdot \frac{\sin 3x}{3x} \to 6$. The cancellation of one $\sin 3x$ is legal because $\sin 3x \ne 0$ for $x$ near $0$, $x \ne 0$. Writing $1 - \cos 6x = 2\sin^2 6x$ (the angle not halved) gives $\frac{2\sin^2 6x}{x \sin 3x}$, which leads nowhere.

c) Pythagoras: $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. For $x$ near $0$, $x \ne 0$, $\cos x \ne 1$, so $\frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} = \frac{1}{1 + \cos x} \to \frac{1}{2}$. No $\frac{\sin \theta}{\theta}$ is needed at all: the identity alone removes the $\frac{0}{0}$. Recognizing a difference of squares hidden in $\sin^2 x$ is exactly the algebra the course's tutorials drill.

d) Half angle: $1 - \cos 2x = 2\sin^2 x$, so $\sqrt{1 - \cos 2x} = \sqrt 2 \sqrt{\sin^2 x} = \sqrt 2\,|\sin x|$. For $0 < x < \pi$, $\sin x > 0$ and the quotient is $\sqrt 2 \frac{\sin x}{x} \to \sqrt 2$. For $-\pi < x < 0$, $\sin x < 0$, so $|\sin x| = -\sin x$ and the quotient is $-\sqrt 2 \frac{\sin x}{x} \to -\sqrt 2$. The one-sided limits differ: no two-sided limit. The student who writes $\sqrt{\sin^2 x} = \sin x$ finds $\sqrt 2$ on both sides and loses the question.

e) Divide numerator and denominator by $x$: $\frac{1 + \frac{\sin x}{x}}{1 - 3\frac{\sin x}{x}}$. The numerator tends to $2$, the denominator to $1 - 3 = -2 \ne 0$, so the quotient rule gives $\frac{2}{-2} = -1$. Replacing $\sin x$ by $x$ “because they are close” gives $\frac{2x}{-2x} = -1$ here by luck, and earns nothing: the division by $x$ followed by the theorem is the justification.