Corrected exercises: indeterminate forms and L'Hospital's rule (MATH 140)

a) As $x \to 0$, $e^{3x} - 1 \to e^0 - 1 = 0$ and $\sin 2x \to 0$: the form is $\frac{0}{0}$, and both functions are differentiable, with $(\sin 2x)' = 2\cos 2x \ne 0$ near $0$. By L'Hospital's rule, with the chain rule on each inner function $3x$ and $2x$: $\lim_{x\to 0} \frac{3e^{3x}}{2\cos 2x} = \frac{3 \cdot 1}{2 \cdot 1} = \frac{3}{2}$. The figure explains why: near $0$ each curve is almost its tangent line, $e^{3x} - 1 \approx 3x$ and $\sin 2x \approx 2x$, so the quotient is almost $\frac{3x}{2x} = \frac{3}{2}$. L'Hospital's rule is exactly that: when both functions vanish at $a$, their quotient tends to the quotient of their SLOPES at $a$. Writing the form $\frac{0}{0}$ before differentiating is the first method mark of every question of this chapter.

b) At $x = 1$, $\ln 1 = 0$ and $1^2 - 1 = 0$: the form is $\frac{0}{0}$. By L'Hospital's rule, $\lim_{x\to 1} \frac{1/x}{2x} = \frac{1}{2}$. The new quotient is continuous at $1$, so the limit is read by substitution and the computation stops after one round. Factoring does not work here, since $\ln x$ has no factor $x - 1$ to cancel: this is the kind of $\frac{0}{0}$ for which the rule was made.

c) At $x = 2$: $8 - 8 = 0$ and $4 + 2 - 6 = 0$, form $\frac{0}{0}$. By L'Hospital's rule, $\lim_{x\to 2} \frac{3x^2}{2x + 1} = \frac{12}{5}$. By factoring: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ and $x^2 + x - 6 = (x - 2)(x + 3)$, so for $x \ne 2$ the quotient is $\frac{x^2 + 2x + 4}{x + 3} \to \frac{12}{5}$. The answers agree, as they must: the rule is a theorem, not a trick. For polynomials the factorization is often just as fast, and it is the method to keep when a question says without L'Hospital's rule.

d) As $x \to \pi^-$, $\sin x \to 0$ but $1 - \cos x \to 1 - (-1) = 2$. The form is $\frac{0}{2}$, which is NOT indeterminate: the limit is $\frac{0}{2} = 0$, by the quotient law, and the question is over. The student skipped the check of the form, and the rule, applied outside its hypotheses, produced $-\infty$, a wrong answer with a correct-looking computation. This is the single most expensive error of the chapter: the whole question is lost, and nothing on the page looks wrong to the student.

e) At $x = 0$: $5^0 - 3^0 = 0$ and the denominator is $0$, form $\frac{0}{0}$. Write $5^x = e^{x\ln 5}$, so $(5^x)' = 5^x \ln 5$ by the chain rule, and likewise $(3^x)' = 3^x \ln 3$. By L'Hospital's rule, $\lim_{x\to 0} \frac{5^x \ln 5 - 3^x \ln 3}{1} = \ln 5 - \ln 3 = \ln\frac{5}{3}$. The answer is exact: $\ln\frac{5}{3}$, not a decimal. A student who writes $(5^x)' = x \cdot 5^{x-1}$ applies the power rule to an exponential and gets $\frac{0}{1}$ at the end, a sign that the derivative was wrong.

a) As $x \to \infty$, $(\ln x)^2 \to \infty$ and $x \to \infty$: form $\frac{\infty}{\infty}$. Round 1, with the chain rule on the square: $\frac{2\ln x \cdot \frac{1}{x}}{1} = \frac{2\ln x}{x}$. Check again: $\frac{\infty}{\infty}$, so the rule may be used once more. Round 2: $\frac{2/x}{1} = \frac{2}{x} \to 0$. The limit is $0$. Each round is licensed by its own form: writing $\frac{\infty}{\infty}$ twice is not a formality, it is the permit for the second derivative quotient.

b) Form $\frac{\infty}{\infty}$. Round 1: $\frac{e^x}{4x^3}$, still $\frac{\infty}{\infty}$. Round 2: $\frac{e^x}{12x^2}$, still $\frac{\infty}{\infty}$. Round 3: $\frac{e^x}{24x}$, still $\frac{\infty}{\infty}$. Round 4: $\frac{e^x}{24} \to \infty$. So $\frac{e^x}{x^4} \to \infty$. Each round lowers the power by one and leaves $e^x$ unchanged; the rule also allows a limit equal to $\infty$. By the growth ranking the answer was predictable, but on an exam the ranking is quoted only if the course allows it: the four rounds are the proof.

c) Both logarithms tend to $\infty$: form $\frac{\infty}{\infty}$. By the rule, with the chain rule on each inner polynomial: $\frac{\frac{3x^2}{x^3 + 1}}{\frac{2x}{x^2 + 5}} = \frac{3x(x^2 + 5)}{2(x^3 + 1)}$. This is now a rational function, and dividing top and bottom by $x^3$ gives $\frac{3(1 + 5/x^2)}{2(1 + 1/x^3)} \to \frac{3}{2}$. Simplify BETWEEN rounds: differentiating $\frac{3x^2}{x^3+1}$ again with the quotient rule would be a waste of ten lines. The answer is the ratio of the degrees, because $\ln(x^3 + 1)$ behaves like $3\ln x$ and $\ln(x^2 + 5)$ like $2\ln x$.

d) As $x \to 0^+$, $\sin x \to 0^+$, so $\ln(\sin x) \to -\infty$, and $\ln x \to -\infty$: form $\frac{-\infty}{-\infty}$, covered by the rule. Round 1: $\frac{\cos x / \sin x}{1/x} = \frac{x \cos x}{\sin x}$. This is of the form $\frac{0}{0}$; rather than a second round, split it: $\frac{x}{\sin x} \cdot \cos x$, and $\frac{x}{\sin x} \to 1$ (one round of the rule: $\frac{1}{\cos x} \to 1$). So the limit is $1 \cdot 1 = 1$. The logarithm flattens everything: $\sin x$ and $x$ differ, but their logarithms have the same size near $0$.

e) Since $e < 3$, $e^3 < 27$. Since $e > 2.7$, $e^5 > 2.7^5$, and $2.7^2 = 7.29$ and $2.7^4 = 7.29^2 > 53$, so $2.7^5 > 53 \times 2.7 > 143 > 125$. So at $x = 3$ the cubic is ahead, $x^3 > e^x$, and at $x = 5$ the exponential has passed it, as the figure shows: after a first crossing near $x = 2$, the curves cross again between $4$ and $5$ (at $4$ the cubic still leads, $e^4 < 2.72^4 < 64$), and never after that. For the limit: form $\frac{\infty}{\infty}$; three rounds give $\frac{3x^2}{e^x}$, $\frac{6x}{e^x}$, $\frac{6}{e^x} \to 0$, the form $\frac{\infty}{\infty}$ being checked before each. No contradiction: a limit at infinity says nothing about $x = 3$, it describes what happens for ALL $x$ large enough. Concluding from a table of small values that $x^3$ grows faster is the error the figure is designed to provoke.

a) Form $0 \cdot (-\infty)$. First way: $\frac{\sin x}{1/\ln x}$, of the form $\frac{0}{0}$; the rule gives $\frac{\cos x}{-\frac{1}{x(\ln x)^2}} = -x(\ln x)^2 \cos x$, a product of the same kind as before and harder. Abandon it. Second way: $\frac{\ln x}{1/\sin x} = \frac{\ln x}{\csc x}$, of the form $\frac{-\infty}{\infty}$; the rule gives $\frac{1/x}{-\csc x \cot x} = -\frac{\sin x}{x} \cdot \tan x$, and since $\frac{\sin x}{x} \to 1$ and $\tan x \to 0$, the limit is $-1 \cdot 0 = 0$. The rule of thumb: keep the LOGARITHM upstairs, because its derivative $\frac{1}{x}$ is simpler than it, while $\frac{1}{\ln x}$ has an uglier derivative.

b) Form $(-\infty) \cdot 0$. Write $x e^x = \frac{x}{e^{-x}}$: as $x \to -\infty$, $x \to -\infty$ and $e^{-x} \to \infty$, form $\frac{-\infty}{\infty}$. By the rule, $\frac{1}{-e^{-x}} = -e^x \to 0$. So $x e^x \to 0$, from below since $x e^x < 0$ for $x < 0$. The other choice, $\frac{e^x}{1/x}$, is of the form $\frac{0}{0}$ and the rule turns it into $\frac{e^x}{-1/x^2} = -x^2 e^x$: a worse product. Rewriting is a choice, and a round that makes things worse is a signal to go back, not to insist.

c) As $x \to 1^+$, both $\frac{1}{\ln x}$ and $\frac{1}{x - 1}$ tend to $+\infty$: form $\infty - \infty$. Common denominator: $\frac{x - 1 - \ln x}{(x - 1)\ln x}$, of the form $\frac{0}{0}$. Round 1: $\frac{1 - \frac{1}{x}}{\ln x + \frac{x - 1}{x}}$; multiply top and bottom by $x$: $\frac{x - 1}{x\ln x + x - 1}$, still $\frac{0}{0}$. Round 2: $\frac{1}{\ln x + 1 + 1} \to \frac{1}{2}$. The difference of two quantities that both blow up is $\frac{1}{2}$, not $0$: they blow up at the same rate, and the rule measures what is left over. Clearing the small fractions between rounds is what keeps round 2 short.

d) As $x \to (\pi/2)^-$, $\sec x \to \infty$ and $\tan x \to \infty$: form $\infty - \infty$. Common denominator, since both have $\cos x$ downstairs: $\sec x - \tan x = \frac{1 - \sin x}{\cos x}$, of the form $\frac{0}{0}$. By the rule, $\frac{-\cos x}{-\sin x} = \frac{\cos x}{\sin x} \to \frac{0}{1} = 0$. The last quotient is NOT indeterminate, so the computation stops there. Answer: $0$.

e) Form $\infty - \infty$. Here no derivative is needed: by the law of logarithms, $\ln(2x + 1) - \ln(x + 3) = \ln\frac{2x + 1}{x + 3}$. Inside, $\frac{2x + 1}{x + 3} = \frac{2 + 1/x}{1 + 3/x} \to 2$, and $\ln$ is continuous at $2$, so the limit is $\ln 2$. Turning the difference into a quotient is the whole method for $\infty - \infty$; sometimes the quotient is then easy, and the rule is not needed at all. Answering $0$ because the two logarithms both grow without bound confuses a form with a value.

a) Form $\frac{0}{0}$, since $\tan 0 - 0 = 0$. Round 1: $\frac{\sec^2 x - 1}{3x^2} = \frac{\tan^2 x}{3x^2}$, by the identity $\sec^2 x - 1 = \tan^2 x$. Still $\frac{0}{0}$, but a second derivative of $\sec^2 x$ is not needed: $\frac{\tan^2 x}{3x^2} = \frac{1}{3}\left(\frac{\tan x}{x}\right)^2$, and $\frac{\tan x}{x}$ is $\frac{0}{0}$ with, by the rule, $\frac{\sec^2 x}{1} \to 1$. So the limit is $\frac{1}{3} \cdot 1^2 = \frac{1}{3}$. The figure agrees: the curve is even, rises from about $0.33$ near $0$ to about $0.79$ at $x = \pm 1.2$, and the hole sits at $\left(0, \frac{1}{3}\right)$. Three blind rounds would work too, with $2\sec^2 x \tan x$ and then its derivative: the identity saves two error-prone lines.

b) At $0$: $1 + 1 - 2 = 0$ over $0$, form $\frac{0}{0}$. Round 1, chain rule on $e^{-x}$: $\frac{e^x - e^{-x}}{2x}$, form $\frac{0}{0}$ again since $e^0 - e^0 = 0$. Round 2: $\frac{e^x + e^{-x}}{2} \to \frac{1 + 1}{2} = 1$. The sign in round 1 is the trap: $(e^{-x})' = -e^{-x}$, and a student who forgets the minus finds $\frac{e^x + e^{-x}}{2x} \to \frac{2}{0}$, an infinite limit for a function that is visibly bounded near $0$.

c) The first round is legitimate: at $0$ the form is $\frac{1 - 1 - 0}{0} = \frac{0}{0}$, and it gives $\frac{e^x - 1}{2x + 1}$. But now the numerator tends to $0$ and the denominator to $1$: the form is $\frac{0}{1}$, NOT indeterminate, and the limit is $0$. The second round was not allowed: its hypothesis, a form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, is false, and it produced $\frac{1}{2}$, a wrong answer. Check: for $x = 0.01$, $e^x - 1 - x$ is about $\frac{x^2}{2} = 0.00005$ while $x^2 + x \approx 0.01$, a quotient near $0.005$, close to $0$. Applying the rule once too often costs the whole question, and it is invisible unless the form is written before EVERY round.

d) Form $\frac{0}{0}$. Round 1: $\frac{1 - \cos x}{1 - \sec^2 x} = \frac{1 - \cos x}{-\tan^2 x}$, still $\frac{0}{0}$. Simplify instead of differentiating: $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$ and $\sin^2 x = (1 - \cos x)(1 + \cos x)$, so for $x$ near $0$, $x \ne 0$, the quotient is $-\frac{(1 - \cos x)\cos^2 x}{(1 - \cos x)(1 + \cos x)} = -\frac{\cos^2 x}{1 + \cos x} \to -\frac{1}{2}$. The factor $1 - \cos x$ that caused the $\frac{0}{0}$ cancels. Blind rounds would reach the same $-\frac{1}{2}$ after two more derivatives of $\sec^2 x$; the algebra is shorter and harder to get wrong.

a) As $x \to 0^+$ the base $x \to 0^+$ and the exponent $\sin x \to 0$: form $0^0$. Let $y = x^{\sin x}$, so $\ln y = \sin x \ln x$, of the form $0 \cdot (-\infty)$. By Exercise 3 a), rewritten as $\frac{\ln x}{\csc x}$ and one round of the rule, $\ln y \to 0$. The exponential is continuous at $0$, so $y = e^{\ln y} \to e^0 = 1$. The answer is $1$ HERE because the logarithm tends to $0$, not because $0^0$ is $1$: part e) shows the difference.

b) As $x \to 0$, $\cos x \to 1$ and $\frac{1}{x^2} \to \infty$: form $1^\infty$. With $y = (\cos x)^{1/x^2}$ (defined for $|x| < \frac{\pi}{2}$, where $\cos x > 0$), $\ln y = \frac{\ln(\cos x)}{x^2}$, of the form $\frac{0}{0}$. Round 1: $\frac{-\sin x/\cos x}{2x} = -\frac{1}{2} \cdot \frac{\tan x}{x} \to -\frac{1}{2}$, since $\frac{\tan x}{x} \to 1$ (Exercise 4 a). So $y \to e^{-1/2} = \frac{1}{\sqrt e}$, about $0.61$ with $e \approx 2.72$. This is the height of the dashed line, and the figure shows the curve closing in on it from both sides, with a hole at $\left(0, e^{-1/2}\right)$. The frequent error is to stop at $\ln y \to -\frac{1}{2}$ and answer $-\frac{1}{2}$: a power of a positive number is never negative, and the last line must exponentiate.

c) As $x \to \infty$, the base $e^x + x \to \infty$ and the exponent $\frac{1}{x} \to 0$: form $\infty^0$. $\ln y = \frac{\ln(e^x + x)}{x}$, form $\frac{\infty}{\infty}$. Round 1: $\frac{e^x + 1}{e^x + x}$, form $\frac{\infty}{\infty}$ again. A second round gives $\frac{e^x}{e^x + 1}$ and a third $\frac{e^x}{e^x} = 1$; faster, divide top and bottom of $\frac{e^x + 1}{e^x + x}$ by $e^x$: $\frac{1 + e^{-x}}{1 + x e^{-x}} \to \frac{1 + 0}{1 + 0} = 1$, since $x e^{-x} = \frac{x}{e^x} \to 0$. So $\ln y \to 1$ and $y \to e$. The term $x$ is negligible next to $e^x$, and $(e^x)^{1/x} = e$ exactly: the limit says the $x$ does not matter.

d) As $x \to 0^+$, $1 + \sin 4x \to 1$ and $\cot x \to \infty$: form $1^\infty$. $\ln y = \cot x \ln(1 + \sin 4x) = \frac{\ln(1 + \sin 4x)}{\tan x}$, form $\frac{0}{0}$. Round 1, chain rule twice on the numerator: $\frac{\frac{4\cos 4x}{1 + \sin 4x}}{\sec^2 x} \to \frac{4 \cdot 1 / 1}{1} = 4$. So $y \to e^4$. Writing $\cot x = \frac{1}{\tan x}$ is the move that turns the product into a quotient; the inner function $4x$ must bring its factor $4$, and it is exactly that $4$ which ends up in the exponent of the answer.

e) For $0 < x < 1$, $\ln x \ne 0$ and $x^{1/\ln x} = e^{\frac{1}{\ln x} \cdot \ln x} = e^1 = e$. As $x \to 0^+$ the base tends to $0$ and the exponent $\frac{1}{\ln x}$ tends to $0^-$: the form is $0^0$, yet the function is the CONSTANT $e$, and its limit is $e$, not $1$. The student confuses the number $0^0$, which some texts define as $1$ for convenience, with the FORM $0^0$, which is a statement about two functions that both tend to $0$ and says nothing about their race. The same holds for $1^\infty$ (part b gave $e^{-1/2}$) and $\infty^0$ (part c gave $e$).