Corrected exercises: limits and the limit laws (MATH 140)

As $x$ approaches $-2$ from the left, the curve runs down to the height $1$, where the empty dot sits; from the right it leaves from the same empty dot. So $\lim_{x\to -2^-} f(x) = 1$ and $\lim_{x\to -2^+} f(x) = 1$, and since the two one-sided limits are equal, $\lim_{x\to -2} f(x) = 1$. The full dot at height $3$ gives the value: $f(-2) = 3$. The limit and the value are two different numbers, and nothing is wrong with that: the limit reads the curve AROUND $-2$, the value reads the single isolated dot. Answering $3$ for the limit because the dot is full is the most common error of the chapter.

From the left the curve climbs to the empty dot at height $2$: $\lim_{x\to 1^-} f(x) = 2$. From the right the curve starts at the full dot at height $-1$: $\lim_{x\to 1^+} f(x) = -1$. The one-sided limits exist but are different, so $\lim_{x\to 1} f(x)$ does not exist. The value is given by the full dot: $f(1) = -1$. Note that $f(1)$ equals the limit from the right; this does not rescue the two-sided limit, which requires the left side to agree as well. Writing $\lim_{x\to 1} f(x) = -1$ because $f(1) = -1$ is wrong twice: it uses the value, and it forgets the left side.

Near $4$ the graph is made of two straight pieces that meet at the point $(4, 2)$, one rising, one falling. From the left, $f(x) \to 2$; from the right, $f(x) \to 2$. So $\lim_{x\to 4} f(x) = 2$, and here the value agrees: $f(4) = 2$. The corner changes the DIRECTION of the graph, not its HEIGHT, and a limit only asks about heights. A corner will matter later in the course, for the slope; for the limit it is invisible.

As $x$ increases to $6$ the last piece comes down to the full dot $(6, 0)$, so $\lim_{x\to 6^-} f(x) = 0$. Only the left-hand limit makes sense at $6$, because $f$ is not defined to the right of $6$. Inside $(-4, 6)$, the graph is an unbroken curve except at $-2$ and $1$. At $a = 1$ the one-sided limits differ, so the limit fails to exist there and nowhere else. At $a = -2$ the limit exists, equal to $1$, but differs from $f(-2) = 3$. At every other $a$, including the corner $a = 4$, the limit exists and equals $f(a)$.

The product law cannot be used at $1$, since $\lim_{x\to 1} f(x)$ does not exist; work on each side instead. From the left, $f(x) \to 2$, so $(f(x) - 2)(f(x) + 1) \to (2 - 2)(2 + 1) = 0$. From the right, $f(x) \to -1$, so the product tends to $(-1 - 2)(-1 + 1) = 0$. The two sides agree, so the limit EXISTS and equals $0$, although $f$ itself has no limit at $1$: each side kills a different factor. At $-2$, the power law applies because $\lim_{x\to -2} f(x) = 1$ exists: $\lim_{x\to -2} [f(x)]^2 = 1^2 = 1$, whereas $[f(-2)]^2 = 9$. The limit of the square is the square of the limit, never the square of the value.

Every piece has a finite limit, so the laws apply. Difference and constant multiple: $2 \cdot 4 - 3 \cdot (-2) = 14$. For the quotient, first check the permit: $\lim (f(x) + g(x)) = 4 - 2 = 2 \ne 0$, so the quotient law applies, with the product law on top: $\frac{4 \cdot (-2)}{2} = -4$. Root law: $\lim (f(x) - 2g(x)) = 4 + 4 = 8 > 0$, so the limit is $\sqrt 8 = 2\sqrt 2$. For $\frac{g}{h}$ the quotient law has NO permit, since $\lim h(x) = 0$. Suppose the quotient had a finite limit $M$. Then $g(x) = \frac{g(x)}{h(x)} \cdot h(x)$ is a product of two functions with limits, and the product law gives $\lim g(x) = M \cdot 0 = 0$, contradicting $\lim g(x) = -2$. So $\frac{g(x)}{h(x)}$ has no finite limit. This argument, the product law used BACKWARDS, is the tool of c), of e) and of Exercise 9.

Denominator first: by the sum, power and constant laws, $\lim_{x\to -1} (x^2 + 3) = (-1)^2 + 3 = 4 \ne 0$, which gives the quotient law its permit. Numerator: $\lim_{x\to -1} x^3 + 2\lim_{x\to -1} x^2 - \lim_{x\to -1} 5 = -1 + 2 - 5 = -4$, by the sum, difference, constant multiple and power laws. Quotient law: the limit is $\frac{-4}{4} = -1$. The one-line route is the direct substitution property: the function is rational and $-1$ is in its domain, since $x^2 + 3$ never vanishes, so the limit is the value at $-1$. That property is itself the limit laws applied once and for all, and it is the ONLY situation in which plugging in is a justified method.

The hypothesis cannot be split by the quotient law, because the denominator $x - 2$ tends to $0$. Write instead, for $x \ne 2$, $f(x) = \frac{f(x) - 5}{x - 2} \cdot (x - 2) + 5$. Each piece now has a limit: the fraction tends to $3$, and $x - 2$ tends to $0$. By the product and sum laws, $\lim_{x\to 2} f(x) = 3 \cdot 0 + 5 = 5$. As for $f(2)$, the hypothesis only involves $x \ne 2$, so it says NOTHING about $f(2)$: $f$ could even be undefined at $2$. The trap is to conclude $f(2) = 5$ because otherwise the fraction would blow up: the limit never looks at $x = 2$.

For $x > 0$, $u(x) = \frac{x}{x} = 1$, and for $x < 0$, $u(x) = \frac{x}{-x} = -1$. So $\lim_{x\to 0^+} u(x) = 1$ and $\lim_{x\to 0^-} u(x) = -1$: different, so $u$ has no limit at $0$, and neither has $v = -u$. Yet $u(x) + v(x) = 0$ for every $x \ne 0$, so $\lim (u + v) = 0$, and $u(x)v(x) = -u(x)^2 = -1$ for every $x \ne 0$, so $\lim uv = -1$. The laws say: IF the limits of the pieces exist, THEN the limit of the combination exists and is computed piece by piece. They do not say the converse, and a combination may have a limit while its pieces have none. When the pieces have no limit, simplify the combination first, then take its limit.

The same backwards product law as in c). For $x \ne 0$, $F(x) = \frac{F(x)}{x^2} \cdot x^2$, a product of a function with limit $5$ and a function with limit $0$: $\lim F(x) = 5 \cdot 0 = 0$. Likewise $\frac{F(x)}{x} = \frac{F(x)}{x^2} \cdot x \to 5 \cdot 0 = 0$. A frequent wrong answer is $\lim \frac{F(x)}{x} = 5$, obtained by cancelling an $x$ in the hypothesis. Always multiply the known quotient by what is missing, then apply the product law, whose permit is now in order. Example: $F(x) = 5x^2 + x^3$ satisfies the hypothesis, and indeed $F(x) \to 0$ and $\frac{F(x)}{x} = 5x + x^2 \to 0$.

Substitution gives $\frac{9 - 3 - 6}{9 - 9} = \frac{0}{0}$: write the form, it tells you that $x + 3$ divides both. $x^2 + x - 6 = (x + 3)(x - 2)$ and $x^2 + 3x = x(x + 3)$. For $x \ne -3$, $\frac{(x + 3)(x - 2)}{x(x + 3)} = \frac{x - 2}{x}$, a rational function with $-3$ in its domain, so by direct substitution the limit is $\frac{-5}{-3} = \frac{5}{3}$. The sign is where marks go: $\frac{-5}{-3}$ is positive. Check the factorization by expanding $(x + 3)(x - 2) = x^2 + x - 6$ before cancelling.

$\frac{8 - 8}{4 - 10 + 6} = \frac{0}{0}$. The numerator is a difference of cubes, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$. The denominator is $(x - 2)(x - 3)$. For $x \ne 2$ the quotient is $\frac{x^2 + 2x + 4}{x - 3}$, and substitution gives $\frac{4 + 4 + 4}{-1} = -12$. Writing $x^3 - 8 = (x - 2)^3$ is the classic slip. Test it at $x = 0$: $-8$ on both sides, which agrees by accident; at $x = 1$ the left gives $-7$ and the right $-1$. One test value is not enough to check a factorization, two are much safer.

$\frac{1 - 1}{1 - 1} = \frac{0}{0}$. $t^4 - 1 = (t^2 - 1)(t^2 + 1) = (t - 1)(t + 1)(t^2 + 1)$, and $t^3 - 1 = (t - 1)(t^2 + t + 1)$. For $t \ne 1$ the quotient is $\frac{(t + 1)(t^2 + 1)}{t^2 + t + 1}$, which tends to $\frac{2 \cdot 2}{3} = \frac{4}{3}$ by direct substitution. The variable is called $t$, and the method does not care: the factor to extract is always the variable minus $a$.

$1 - 3 + 2 = 0$, so $x - 1$ divides the numerator; long or synthetic division gives $x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)^2(x + 2)$, since $x^2 + x - 2$ vanishes at $1$ again. First limit: for $x \ne 1$, $\frac{(x - 1)^2(x + 2)}{(x - 1)(x + 1)} = \frac{(x - 1)(x + 2)}{x + 1} \to \frac{0 \cdot 3}{2} = 0$. Second limit: $\frac{(x - 1)^2(x + 2)}{(x - 1)^2} = x + 2 \to 3$. Two quotients of the same form $\frac{0}{0}$ at the same point give $0$ and $3$: the form carries no value. What decides is how many factors $x - 1$ each side contains, and after the first cancellation the new expression must be read again, since it need not be of the form $\frac{0}{0}$ any more.

The equality of functions is false: the left side is not defined at $x = 3$ while the right side is, so they are two different functions. The correct statement is $\frac{x^2 - x - 6}{x - 3} = \frac{(x - 3)(x + 2)}{x - 3} = x + 2$ FOR $x \ne 3$. The limit is still right because a limit as $x \to 3$ only uses values with $x \ne 3$, where the two functions agree, so the rule of the introduction gives $\lim_{x\to 3} \frac{x^2 - x - 6}{x - 3} = \lim_{x\to 3} (x + 2) = 5$. A marker takes a rigour mark for the missing condition $x \ne 3$: it is precisely the reason the cancellation is allowed.

Substitution gives $\frac{3 - 3}{0} = \frac{0}{0}$. Multiply top and bottom by the conjugate $\sqrt{x + 4} + 3$: the numerator becomes $(x + 4) - 9 = x - 5$, and the quotient is $\frac{x - 5}{(x - 5)(\sqrt{x + 4} + 3)} = \frac{1}{\sqrt{x + 4} + 3}$ for $x \ne 5$. By the quotient and root laws, the limit is $\frac{1}{3 + 3} = \frac{1}{6}$. Multiplying only the numerator by the conjugate changes the value of the expression, and every mark after that line is lost: the conjugate goes on top AND bottom, which is multiplying by $1$.

$\frac{1 - 1}{0} = \frac{0}{0}$. The conjugate of a DIFFERENCE of two roots is their sum: $\left(\sqrt{1 + x} - \sqrt{1 - x}\right)\left(\sqrt{1 + x} + \sqrt{1 - x}\right) = (1 + x) - (1 - x) = 2x$. So for $x \ne 0$ the quotient equals $\frac{2x}{x\left(\sqrt{1 + x} + \sqrt{1 - x}\right)} = \frac{2}{\sqrt{1 + x} + \sqrt{1 - x}}$, which tends to $\frac{2}{1 + 1} = 1$. The minus signs are the trap: $(1 + x) - (1 - x)$ is $2x$, not $0$ and not $2$.

$\frac{0}{2 - 2} = \frac{0}{0}$. Two routes. Conjugate of the denominator: multiply by $2 + \sqrt x$, the bottom becomes $4 - x = -(x - 4)$, and the quotient is $\frac{(x - 4)(2 + \sqrt x)}{-(x - 4)} = -(2 + \sqrt x)$ for $x \ne 4$. Or factor directly, $x - 4 = (\sqrt x - 2)(\sqrt x + 2)$ and $2 - \sqrt x = -(\sqrt x - 2)$. Either way the limit is $-(2 + 2) = -4$. The minus sign comes from $4 - x = -(x - 4)$; losing it gives $4$, and the sign check is immediate: for $x$ slightly above $4$ the top is positive and the bottom negative, so the answer must be negative.

As $x \to 2$, the denominators $x - 2$ and $x^2 - 4$ tend to $0$ while the numerators do not, so by the argument of Exercise 2 a) neither term has a finite limit. The difference law needs both limits to exist: it has no permit. Combine first, over the common denominator $x^2 - 4 = (x - 2)(x + 2)$: $\frac{(x + 2) - 4}{(x - 2)(x + 2)} = \frac{x - 2}{(x - 2)(x + 2)} = \frac{1}{x + 2}$ for $x \ne 2$. The limit is $\frac{1}{4}$. Two terms without limits can have a difference with a limit, exactly as in Exercise 2 d); the common denominator is what makes it visible.

$\frac{2 - 2}{0} = \frac{0}{0}$. The square root conjugate would produce $\sqrt[3]{x}^2 - 4$, still a root. The identity that removes a CUBE root is the difference of cubes: with $a = \sqrt[3]{x}$ and $b = 2$, $x - 8 = a^3 - b^3 = (a - 2)(a^2 + 2a + 4)$. So for $x \ne 8$, $\frac{a - 2}{(a - 2)(a^2 + 2a + 4)} = \frac{1}{\sqrt[3]{x^2} + 2\sqrt[3]{x} + 4}$, which tends to $\frac{1}{4 + 4 + 4} = \frac{1}{12}$. The conjugate is not a magic trick for square roots, it is the identity $a^2 - b^2 = (a - b)(a + b)$; the index of the root tells you which identity to use.

Substitution gives $\frac{0}{16 - 20 + 4} = \frac{0}{0}$, and the inside $x + 4$ vanishes at $-4$: split. The denominator factors as $(x + 1)(x + 4)$. For $x > -4$, $|x + 4| = x + 4$ and the quotient is $\frac{1}{x + 1} \to \frac{1}{-3} = -\frac{1}{3}$. For $x < -4$, $|x + 4| = -(x + 4)$ and the quotient is $-\frac{1}{x + 1} \to \frac{1}{3}$. The one-sided limits differ, so the limit does not exist. Computing only one side, usually the right one, and announcing $-\frac{1}{3}$ loses the whole question.