Corrected exercises: Indeterminate forms and L'Hôpital's Rule (MATH 203)

a) At $x = 3$: $3^2 - 9 = 0$ and $\ln(3 - 2) = \ln 1 = 0$, so the form is $\frac{0}{0}$ and the rule applies. The derivative of the top is $2x$; the derivative of the bottom is $\frac{1}{x - 2}$, by the chain rule with inner function $x - 2$. The new quotient is a complex fraction, and it is cleared before anything else: $\frac{2x}{\frac{1}{x - 2}} = 2x \cdot \frac{x - 2}{1} = 2x(x - 2)$. Dividing by a fraction is multiplying by its reciprocal; a student who writes $\frac{2x}{x - 2}$ has divided by the wrong thing and finds $\frac{6}{0}$, an infinite limit for a quotient the figure shows to be tame. Now $2x(x - 2) \to 6 \cdot 1 = 6$. The figure explains the number: near $3$, $x^2 - 9 \approx 6(x - 3)$ and $\ln(x - 2) \approx 1 \cdot (x - 3)$, so the quotient is close to $\frac{6(x - 3)}{x - 3} = 6$. When both functions vanish at $a$, their quotient tends to the quotient of their slopes at $a$: that is the whole content of the rule.

b) At $0$: $\arctan 0 = 0$ and $e^0 - 1 = 0$, form $\frac{0}{0}$. By the rule, with the chain rule on the inner function $2x$: $\lim_{x\to 0} \frac{\frac{2}{1 + 4x^2}}{e^x}$. Clear the fraction: $\frac{2}{(1 + 4x^2)e^x}$, which is continuous at $0$, so the limit is $\frac{2}{1 \cdot 1} = 2$. The factor $2$ is the one the inner function brings: forgetting it gives $1$, and nothing on the page looks wrong.

c) At $\frac{\pi}{2}$: $1 - \sin\frac{\pi}{2} = 0$ and $\cos^2\frac{\pi}{2} = 0$, form $\frac{0}{0}$. Round 1, with the chain rule on the square: $\frac{-\cos x}{2\cos x \cdot (-\sin x)} = \frac{-\cos x}{-2\sin x \cos x}$. Substituting now gives $\frac{0}{0}$ again, and a student who starts a second round differentiates a product for nothing. The factor $\cos x$ that causes the $\frac{0}{0}$ appears on both levels: for $x$ near $\frac{\pi}{2}$, $\cos x \ne 0$, and it cancels, leaving $\frac{1}{2\sin x} \to \frac{1}{2}$. The same answer comes without the rule, by the identity $\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)$: the quotient is $\frac{1}{1 + \sin x} \to \frac{1}{2}$. Cancelling a common factor is algebra from the first week of the course, and it is the shortest path in both methods.

d) At $0$ the numerator is $0 + \sin 0 = 0$ but the denominator is $e^0 + 0 = 1$. The form is $\frac{0}{1}$, which is NOT indeterminate, and the quotient law gives the limit $\frac{0}{1} = 0$ at once. The hypothesis $g(a) = 0$ of the rule is false here, so the rule may not be used, and applied anyway it produced $1$, a wrong answer. The error is the missing first line: the form was never written. On an exam this costs the whole question, because the page shows no calculation error at all.

e) At $0$: $3^0 - 1 = 0$ and $2^0 - 1 = 0$, form $\frac{0}{0}$. The derivative of $a^x$ is $a^x \ln a$ (write $a^x = e^{x\ln a}$ and use the chain rule). By the rule, $\lim_{x\to 0} \frac{3^x \ln 3}{2^x \ln 2} = \frac{\ln 3}{\ln 2}$, which is $\log_2 3$ by the change of base formula. With the calculator, $\frac{1.098612\ldots}{0.693147\ldots} \approx 1.5850$. The exact answer is $\frac{\ln 3}{\ln 2}$; the decimal only answers the second half of the question. Writing $(3^x)' = x \cdot 3^{x - 1}$, the power rule applied to an exponential, gives $\frac{0}{0}$ again at $0$ and loops: the variable is in the EXPONENT.

a) As $x \to \infty$, $(\ln x)^3 \to \infty$ and $x^2 \to \infty$: form $\frac{\infty}{\infty}$. Round 1, chain rule on the cube with inner function $\ln x$: $\frac{3(\ln x)^2 \cdot \frac{1}{x}}{2x}$. Clear it: $\frac{3(\ln x)^2}{x \cdot 2x} = \frac{3(\ln x)^2}{2x^2}$. The form is $\frac{\infty}{\infty}$ again. Round 2: $\frac{6\ln x \cdot \frac{1}{x}}{4x} = \frac{3\ln x}{2x^2}$, still $\frac{\infty}{\infty}$. Round 3: $\frac{3 \cdot \frac{1}{x}}{4x} = \frac{3}{4x^2} \to 0$. Each round lowers the power of $\ln x$ by one, and only because the $\frac{1}{x}$ was moved down next to $x$ each time. The frequent error is to read $\frac{\frac{1}{x}}{2x}$ as $\frac{2x}{x}$, flipping the wrong level: the limit then comes out as $\infty$.

b) Form $\frac{\infty}{\infty}$. Round 1, with the factor $\frac{1}{3}$ of the inner function $\frac{x}{3}$: $\frac{\frac{1}{3}e^{x/3}}{2x} = \frac{e^{x/3}}{6x}$, still $\frac{\infty}{\infty}$. Round 2: $\frac{\frac{1}{3}e^{x/3}}{6} = \frac{e^{x/3}}{18} \to \infty$. The limit is $+\infty$: the rule also allows an infinite limit on the right side. The slower exponential $e^{x/3}$ still beats every power of $x$, it only needs larger $x$ to do it.

c) As $x \to \left(\frac{\pi}{2}\right)^-$, $\cos x \to 0^+$ so $\ln(\cos x) \to -\infty$, and $\tan x \to +\infty$: form $\frac{-\infty}{\infty}$. Round 1: the top gives $\frac{-\sin x}{\cos x}$ (chain rule, inner function $\cos x$), the bottom gives $\sec^2 x = \frac{1}{\cos^2 x}$. Clear the complex fraction: $\frac{-\sin x}{\cos x} \cdot \cos^2 x = -\sin x \cos x$. This is a product of continuous functions, and it tends to $-1 \cdot 0 = 0$. The limit is $0$: the tangent blows up faster than the logarithm of the cosine goes down. Left uncleared, $\frac{-\tan x}{\sec^2 x}$ looks like another $\frac{\infty}{\infty}$, and a second round only makes it worse.

d) Form $\frac{\infty}{\infty}$. With the rule: $\frac{6x - \frac{1}{x}}{2x + \frac{4}{x}}$. Multiply top and bottom by $x$ to clear the small fractions: $\frac{6x^2 - 1}{2x^2 + 4}$, and dividing by $x^2$ gives $\frac{6 - 1/x^2}{2 + 4/x^2} \to 3$. Without the rule: divide top and bottom by $x^2$, $\frac{3 - \frac{\ln x}{x^2}}{1 + 4\frac{\ln x}{x^2}}$, and $\frac{\ln x}{x^2} \to 0$ (one round: $\frac{1/x}{2x} = \frac{1}{2x^2} \to 0$), so the limit is $\frac{3}{1} = 3$. The answers agree. The second method is the chapter 5 method, dividing by the dominant term; the rule was only needed to know that $\ln x$ is negligible next to $x^2$.

a) Form $0 \cdot \infty$, since $x \to 0^+$ and $(\ln x)^2 \to +\infty$. First way: $\frac{x}{1/(\ln x)^2}$, form $\frac{0}{0}$; the derivative of $(\ln x)^{-2}$ is $-2(\ln x)^{-3} \cdot \frac{1}{x}$, and the new quotient is $\frac{1}{-2(\ln x)^{-3}/x} = -\frac{x(\ln x)^3}{2}$, a product of the same kind with a HIGHER power of the logarithm. Abandon it. Second way: $x = \frac{1}{x^{-1}}$, so $x(\ln x)^2 = \frac{(\ln x)^2}{x^{-1}}$, form $\frac{\infty}{\infty}$. Round 1: $\frac{2\ln x \cdot x^{-1}}{-x^{-2}} = -2\ln x \cdot x^{-1} \cdot x^{2} = -2x\ln x$, again $0 \cdot (-\infty)$. Rewrite the same way: $-2x\ln x = \frac{-2\ln x}{x^{-1}}$, form $\frac{-\infty}{\infty}$, and round 2 gives $\frac{-2x^{-1}}{-x^{-2}} = 2x \to 0$. The limit is $0$. The rule of thumb: keep the logarithm upstairs, where differentiating simplifies it, and send the power down with its exponent flipped.

b) As $x \to 1^-$, $1 - x \to 0^+$ and $\frac{\pi x}{2} \to \left(\frac{\pi}{2}\right)^-$, so $\tan\frac{\pi x}{2} \to +\infty$: form $0 \cdot \infty$. Since $\tan u = \frac{1}{\cot u}$, write $\frac{1 - x}{\cot(\pi x/2)}$, form $\frac{0}{0}$. The derivative of $\cot u$ is $-\csc^2 u$, times the factor $\frac{\pi}{2}$ of the inner function $u = \frac{\pi x}{2}$. Round 1: $\frac{-1}{-\frac{\pi}{2}\csc^2(\pi x/2)} = \frac{2}{\pi}\sin^2\frac{\pi x}{2} \to \frac{2}{\pi} \cdot 1 = \frac{2}{\pi}$, about $0.6366$. The other choice, $\frac{\tan(\pi x/2)}{(1 - x)^{-1}}$, gives after one round $\frac{\frac{\pi}{2}\sec^2(\pi x/2)}{(1 - x)^{-2}}$, which is worse. The exact answer is $\frac{2}{\pi}$; forgetting the chain factor $\frac{\pi}{2}$ gives $1$.

c) As $x \to 0^+$, $\cot x \to +\infty$ and $\frac{1}{x} \to +\infty$; as $x \to 0^-$, both tend to $-\infty$: form $\infty - \infty$ on each side. Common denominator: $\cot x - \frac{1}{x} = \frac{\cos x}{\sin x} - \frac{1}{x} = \frac{x\cos x - \sin x}{x\sin x}$, form $\frac{0}{0}$. Round 1: the top gives $\cos x - x\sin x - \cos x = -x\sin x$ (product rule), the bottom gives $\sin x + x\cos x$. The quotient $\frac{-x\sin x}{\sin x + x\cos x}$ is still $\frac{0}{0}$; divide top and bottom by $x$ instead of differentiating: $\frac{-\sin x}{\frac{\sin x}{x} + \cos x} \to \frac{0}{1 + 1} = 0$, using $\frac{\sin x}{x} \to 1$ (Thomas 2.4). The limit is $0$. Answering $0$ at the start because both terms are infinite would be the right number for a wrong reason, and part d) shows that the reason fails.

d) As $x \to 1^+$ both terms tend to $+\infty$, and as $x \to 1^-$ both tend to $-\infty$ (since $x^2 - 1 \to 0^-$ and $x - 1 \to 0^-$): form $\infty - \infty$. Common denominator, with $x^2 - 1 = (x - 1)(x + 1)$: $\frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} = \frac{1 - x}{(x - 1)(x + 1)} = \frac{-1}{x + 1}$ for $x \ne 1$. The sign is the algebra trap: $1 - x = -(x - 1)$, and dropping that minus gives $+\frac{1}{2}$. So the limit is $-\frac{1}{2}$, with no derivative at all. The rule would also work on $\frac{1 - x}{x^2 - 1}$, giving $\frac{-1}{2x} \to -\frac{1}{2}$, but the factorization is shorter. The two infinities differ by a quantity that tends to $-\frac{1}{2}$, not $0$.

e) Both terms tend to $\infty$: form $\infty - \infty$. There is no fraction to combine, so factor out the term expected to dominate: $x^{1/3} - \ln x = x^{1/3}\left(1 - \frac{\ln x}{x^{1/3}}\right)$. The quotient $\frac{\ln x}{x^{1/3}}$ is $\frac{\infty}{\infty}$; one round gives $\frac{x^{-1}}{\frac{1}{3}x^{-2/3}} = 3x^{-1 + 2/3} = 3x^{-1/3} = \frac{3}{x^{1/3}} \to 0$. Subtracting the exponents, $-1 - (-\frac{2}{3}) = -\frac{1}{3}$, is the step to write out. So the bracket tends to $1$, the factor $x^{1/3}$ to $\infty$, and the limit is $+\infty$: even a cube root beats the logarithm. Not everywhere, though: at $x = 27$ the root is $3$ while $\ln 27 \approx 3.30$, and the logarithm is ahead between about $x = 6$ and $x = 93$; at $x = 1000$ the root is $10$ against $\ln 1000 \approx 6.91$, and it leads for good. A limit at infinity only speaks of large $x$.

a) At $0$: $\ln 1 - \sin 0 = 0$ over $0$, form $\frac{0}{0}$. Round 1: $\frac{\frac{1}{1 + x} - \cos x}{2x}$, and at $0$ the top is $1 - 1 = 0$: form $\frac{0}{0}$ again. Round 2: the derivative of $(1 + x)^{-1}$ is $-(1 + x)^{-2}$, and that of $-\cos x$ is $+\sin x$, so the quotient is $\frac{-(1 + x)^{-2} + \sin x}{2} \to \frac{-1 + 0}{2} = -\frac{1}{2}$. Writing $\frac{1}{1 + x}$ as $(1 + x)^{-1}$ before differentiating is what keeps the minus sign: a student who differentiates $\frac{1}{1 + x}$ as $\ln(1 + x)$, or forgets the minus of the power rule, answers $+\frac{1}{2}$. The sign is plausible: near $0$, $\ln(1 + x)$ lies below $x$ by about $\frac{x^2}{2}$, while $\sin x$ stays much closer to $x$.

b) At $0$: $1 - \cos 0 = 0$ and $0 \cdot \sin 0 = 0$, form $\frac{0}{0}$. Round 1: the top gives $4\sin 4x$ (chain rule, inner function $4x$), the bottom $\sin 2x + 2x\cos 2x$ (product rule, with the chain factor $2$). At $0$: $\frac{0}{0}$ again. Round 2: the top gives $16\cos 4x$; the bottom gives $2\cos 2x + 2\cos 2x - 4x\sin 2x = 4\cos 2x - 4x\sin 2x$. At $0$: $\frac{16}{4}$, not indeterminate: stop. The limit is $4$. The product rule on $x\sin 2x$ is where the page goes wrong: $(x\sin 2x)' = \sin 2x + 2x\cos 2x$, two terms, each with its chain factor. A student who writes $(x\sin 2x)' = 2\cos 2x$ finds $\frac{0}{2}$ after round 1 and answers $0$.

c) Rounds 1 and 2 are legal: at $0$ the starting quotient is $\frac{0}{0}$, and so is $\frac{2\cos 2x - 2}{2x + 3x^2}$, since $2\cos 0 - 2 = 0$. But the third quotient, $\frac{-4\sin 2x}{2 + 6x}$, has a denominator that tends to $2$: its form is $\frac{0}{2}$, NOT indeterminate, and its limit is $0$. The third round had no permit, and it produced $-\frac{4}{3}$. The correct limit is $0$. A check confirms it: for small $x$, $\sin 2x - 2x$ is about $-\frac{4}{3}x^3$, so the quotient behaves like $\frac{-\frac{4}{3}x^3}{x^2} = -\frac{4}{3}x \to 0$. The $-\frac{4}{3}$ the student found is the limit of $\frac{\sin 2x - 2x}{x^3}$, another question with another denominator. One round too many costs the whole question, and it only shows if the form is written before EVERY round.

d) At $0$: $\arcsin 0 - 0 = 0$ over $0$, form $\frac{0}{0}$. Round 1, with $(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$: $\frac{\frac{1}{\sqrt{1 - x^2}} - 1}{3x^2}$, still $\frac{0}{0}$. A second round would differentiate $(1 - x^2)^{-1/2}$ and bring a new fraction. Algebra instead. Put the top over one denominator: $\frac{1 - \sqrt{1 - x^2}}{3x^2\sqrt{1 - x^2}}$. Multiply top and bottom by the conjugate $1 + \sqrt{1 - x^2}$: the top becomes $1 - (1 - x^2) = x^2$, which cancels with the $x^2$ below, for $x \ne 0$: $\frac{1}{3\sqrt{1 - x^2}\left(1 + \sqrt{1 - x^2}\right)} \to \frac{1}{3 \cdot 1 \cdot 2} = \frac{1}{6}$. The conjugate is the chapter 3 gesture, and it is exactly what makes the factor $x^2$ appear so that it can cancel.

a) As $x \to 0^+$, the base $\sin x \to 0^+$ and the exponent $\tan x \to 0$: form $0^0$. Let $y = (\sin x)^{\tan x}$; then $\ln y = \tan x \ln(\sin x)$, of the form $0 \cdot (-\infty)$. Send $\tan x$ down as its reciprocal: $\ln y = \frac{\ln(\sin x)}{\cot x}$, form $\frac{-\infty}{\infty}$. Round 1: the top gives $\frac{\cos x}{\sin x}$ (chain rule, inner function $\sin x$), the bottom $-\csc^2 x = -\frac{1}{\sin^2 x}$. Clear the fraction of fractions: $\frac{\cos x}{\sin x} \cdot \left(-\sin^2 x\right) = -\sin x\cos x \to 0$. So $\ln y \to 0$ and $y = e^{\ln y} \to e^0 = 1$. Flipping the wrong level, $-\frac{\cos x}{\sin^3 x}$, would give $-\infty$ and the answer $0$. The answer is $1$ because the logarithm tends to $0$, not because $0^0$ is $1$: part e) shows a base tending to $0$ with another ending.

b) As $x \to 1$ the base tends to $1$, and the exponent $\frac{3}{x - 1}$ tends to $+\infty$ from the right and $-\infty$ from the left: form $1^{\pm\infty}$, NOT $1$. $\ln y = \frac{3}{x - 1}\ln x = \frac{3\ln x}{x - 1}$, form $\frac{0}{0}$ on both sides. Round 1: $\frac{3/x}{1} \to 3$. So $\ln y \to 3$ and $y \to e^3 \approx 20.09$, from both sides. Calculator check: $1.001^{3000} = e^{3000\ln 1.001} \approx 20.06$, close to $20.09$ and far from $1$. The check is done at a MODERATE value: at $x = 1 + 10^{-12}$ a calculator rounds the base to $1$ and displays $1$, which proves nothing.

c) As $x \to \infty$ the base $\ln x \to \infty$ and the exponent $\frac{1}{x} \to 0$: form $\infty^0$. $\ln y = \frac{\ln(\ln x)}{x}$, form $\frac{\infty}{\infty}$. Round 1: the derivative of $\ln(\ln x)$ is $\frac{1}{\ln x} \cdot \frac{1}{x}$ (chain rule, inner function $\ln x$, named), so the quotient is $\frac{1}{x\ln x} \to 0$. Hence $\ln y \to 0$ and $y \to e^0 = 1$. The logarithm grows, but taking its $x$-th root crushes it.