Corrected exercises: L'Hôpital's rule and indeterminate forms, 201-SN3-RE (201-NYB)

Form $\frac{0}{0}$, since $\tan 0=\sin 0=0$. One round: $\lim_{x\to 0}\frac{3\sec^{2}(3x)}{2\cos(2x)}=\frac{3\times 1}{2\times 1}=\frac{3}{2}$. Pitfall: differentiating the fraction with the quotient rule; L'Hôpital's rule differentiates the numerator and the denominator each on its own.

Form $\frac{0}{0}$, since $e^{0}+e^{0}-2=0$. First round: $\frac{e^{x}-e^{-x}}{2x}$, still $\frac{0}{0}$. Second round: $\frac{e^{x}+e^{-x}}{2}\to\frac{1+1}{2}=1$. Pitfall: $(e^{-x})'=-e^{-x}$; forgetting this sign gives, at the first round, $\frac{e^{x}+e^{-x}}{2x}$, of the form $\frac{2}{0}$, and an infinite limit that has nothing to do with the answer.

At $x=2$: $2^{3}-8=0$ and $\ln 1=0$, form $\frac{0}{0}$. One round: $\lim_{x\to 2}\frac{3x^{2}}{\frac{1}{x-1}}=\lim_{x\to 2}3x^{2}(x-1)=12\times 1=12$. The rule holds at any point $a$, not only at $0$: substitute $x=2$ after differentiating.

Form $\frac{\infty}{\infty}$. One round: $\frac{\frac{1}{2\sqrt{x}}}{\frac{2}{x}}=\frac{x}{4\sqrt{x}}=\frac{\sqrt{x}}{4}\to+\infty$. Even the logarithm of a square, $\ln(x^{2})=2\ln x$, is beaten by a square root. Pitfall: $(\ln(x^{2}))'=\frac{2x}{x^{2}}=\frac{2}{x}$, and not $\frac{1}{x^{2}}$.

Form $\frac{\infty}{\infty}$. Three rounds: $\frac{e^{x}+2x}{3e^{x}-1}$, then $\frac{e^{x}+2}{3e^{x}}$, then $\frac{e^{x}}{3e^{x}}=\frac{1}{3}$. Shorter: divide the top and the bottom by $e^{x}$, $\frac{1+x^{2}e^{-x}}{3-xe^{-x}}\to\frac{1+0}{3-0}=\frac{1}{3}$, since the exponential beats every power. Pitfall: stopping at the second round and reading $\frac{e^{x}+2}{3e^{x}}$ as “$\frac{\infty}{\infty}$, so $1$”.

Form $\frac{0}{0}$. One round, then SIMPLIFY: $\frac{\frac{1}{1+x^{2}}-1}{3x^{2}}=\frac{-x^{2}}{3x^{2}(1+x^{2})}=\frac{-1}{3(1+x^{2})}\to-\frac{1}{3}$. Simplifying after the first round saves two differentiations of a quotient.

Form $\frac{0}{0}$, three rounds: $\frac{2-2\cos(2x)}{3x^{2}}$, then $\frac{4\sin(2x)}{6x}$, then $\frac{8\cos(2x)}{6}\to\frac{8}{6}=\frac{4}{3}$. Before each round, check that the form is still $\frac{0}{0}$: it is, at the first two. Pitfall: the factor $2$ of the argument comes back at every differentiation, $(\sin 2x)'=2\cos 2x$ and $(\cos 2x)'=-2\sin 2x$.

Form $\infty\cdot 0$, since $\arctan x\to\frac{\pi}{2}$. Rewrite it as a quotient by sending $x$ to the denominator: $\frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}$, of the form $\frac{0}{0}$. One round: $\frac{-\frac{1}{1+x^{2}}}{-\frac{1}{x^{2}}}=\frac{x^{2}}{1+x^{2}}\to 1$. Pitfall: sending the inverse tangent to the denominator, $\frac{x}{\left(\frac{\pi}{2}-\arctan x\right)^{-1}}$, which also works but makes every derivative heavier.

Form $\infty-\infty$. Common denominator: $\frac{x\ln x-(x-1)}{(x-1)\ln x}$, of the form $\frac{0}{0}$. First round: $\frac{\ln x+1-1}{\ln x+\frac{x-1}{x}}=\frac{\ln x}{\ln x+1-\frac{1}{x}}$, still $\frac{0}{0}$. Second round: $\frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^{2}}}\to\frac{1}{1+1}=\frac{1}{2}$. Pitfall: concluding $\infty-\infty=0$ without rewriting.

Form $1^{\infty}$: the base tends to $1$, the exponent $\frac{2}{1-x}$ to infinity. With $y=x^{\frac{2}{1-x}}$: $\ln y=\frac{2\ln x}{1-x}$, of the form $\frac{0}{0}$, and one round gives $\frac{\frac{2}{x}}{-1}\to-2$. So $y\to e^{-2}\approx 0.135$. Pitfall: answering $1$ (“one to any power”), or answering $-2$, which is the limit of $\ln y$ and not that of $y$.

Form $0^{0}$. $\ln y=x\ln(1-\cos x)=\frac{\ln(1-\cos x)}{\frac{1}{x}}$, of the form $\frac{-\infty}{+\infty}$. One round: $\frac{\frac{\sin x}{1-\cos x}}{-\frac{1}{x^{2}}}=-\frac{x^{2}\sin x}{1-\cos x}$. Since $1-\cos x=\frac{\sin^{2}x}{1+\cos x}$, this equals $-x(1+\cos x)\cdot\frac{x}{\sin x}\to-0\times 2\times 1=0$. So $\ln y\to 0$ and $y\to e^{0}=1$. Pitfall: answering $0$, the limit of $\ln y$.

Form $\infty^{0}$. $\ln y=\frac{\ln(x^{2}+1)}{\ln x}$, of the form $\frac{\infty}{\infty}$. One round: $\frac{\frac{2x}{x^{2}+1}}{\frac{1}{x}}=\frac{2x^{2}}{x^{2}+1}\to 2$. So $y\to e^{2}\approx 7.389$.

$\frac{x^{2}}{x}=x\to 0$; $\frac{x}{x^{3}}=\frac{1}{x^{2}}\to+\infty$; and $\frac{\sin x}{x}\to 1$ (the special limit from the differential calculus course). Three limits of the form $\frac{0}{0}$, three completely different answers: $0$, $+\infty$ and $1$. That is exactly what the comparison shows: the form $\frac{0}{0}$ does not CONTAIN enough information to decide. Everything depends on the relative SPEED at which the numerator and the denominator tend to zero, and that speed does not appear in the symbol. This is why an extra tool is needed, and that is precisely the role of L'Hôpital's rule, which compares those speeds by differentiating.

Because $\frac{1}{0}$ determines the result without ambiguity: if the numerator tends to a nonzero constant and the denominator to zero, the quotient blows up in absolute value. The only thing left to decide is the SIGN, and it is read from the sign of the denominator: $\lim_{x\to 0^{+}}\frac{1}{x}=+\infty$, while $\lim_{x\to 0^{-}}\frac{1}{x}=-\infty$. The form is therefore determinate as to its size, and only a sign study remains. An important practical consequence: applying L'Hôpital's rule to a $\frac{1}{0}$ form is a mistake, since the hypothesis of the rule is not met.

$\lim_{x\to\infty}\frac{\ln x}{x}$: form $\frac{\infty}{\infty}$, INDETERMINATE (it equals 0, as we will see). $\lim_{x\to 0^{+}}\frac{\ln x}{x}$: the numerator tends to $-\infty$ and the denominator to $0^{+}$, so the form is $\frac{-\infty}{0^{+}}$, DETERMINATE, and the limit is $-\infty$: both effects push in the same direction, there is no conflict to settle. $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}$: form $1^{\infty}$, INDETERMINATE (it equals $e$). $\lim_{x\to\infty}\left(2+\frac{1}{x}\right)^{x}$: the base tends to 2, not to 1, so the form is $2^{\infty}$, DETERMINATE, and the limit is $+\infty$. The lesson of part c) is that you must READ the form first of all, and that two almost identical expressions can belong to opposite regimes.

The claim confuses the form and the limit. “Indeterminate” is a judgment on the SYMBOL obtained by direct substitution, not on the existence of the limit. The proof is in part a): $\lim_{x\to 0}\frac{\sin x}{x}=1$ is an indeterminate form whose limit exists perfectly well and equals 1. Correct wording: “direct substitution leads to an indeterminate form, so it does not allow us to conclude; the indeterminate form must be resolved by another method”. The nuance is not rhetorical: it is the difference between giving up on an exercise and starting it.

Complete statement: let $f$ and $g$ be differentiable on an open interval around $a$ (except possibly at $a$), with $g'(x)\neq 0$ on that interval. If $\lim_{x\to a}\frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, AND if $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists (finite or infinite), then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$. The result also holds for $a=\pm\infty$ and for one-sided limits. Application: $\lim_{x\to 0}\frac{e^{x}-1-x}{x^{2}}$ has the form $\frac{0}{0}$; differentiate: $\lim_{x\to 0}\frac{e^{x}-1}{2x}$, still $\frac{0}{0}$; differentiate again: $\lim_{x\to 0}\frac{e^{x}}{2}=\frac{1}{2}$. The limit is $\frac{1}{2}$.

Start: form $\frac{\infty}{\infty}$. First round: $\lim_{x\to\infty}\frac{2x}{e^{x}}$, still $\frac{\infty}{\infty}$. Second round: $\lim_{x\to\infty}\frac{2}{e^{x}}$, and this time the form is $\frac{2}{\infty}$, so DETERMINATE: the limit is $0$. You must stop there: a third round would give $\frac{0}{e^{x}}$, which happens to remain 0 here, but the habit of differentiating without checking the form again ends up costing a lot. The result $\lim_{x\to\infty}\frac{x^{2}}{e^{x}}=0$ is the first case of the hierarchy of growth rates: the exponential beats the power, and each differentiation lowers the degree of the polynomial while $e^{x}$ stays intact. That is the very mechanism of domination.

The mistake is to have applied the rule without checking the hypothesis. At $x=0$, direct substitution gives $\frac{0+1}{0+2}=\frac{1}{2}$: the form is not indeterminate at all, the limit is read off immediately and equals $\frac{1}{2}$. L'Hôpital's rule did not apply, and applying it anyway produced a wrong answer, a limit that would not exist, instead of $\frac{1}{2}$. This is the number one pitfall of the chapter, and it is sneaky because the student's computation is mechanically correct: it is the PREREQUISITES that were missing. Hence the discipline to adopt for the whole semester: write down the form obtained by substitution BEFORE differentiating anything. Three seconds of checking that protect the whole chapter.

The confusion is about the nature of the operation. The quotient rule is used to DIFFERENTIATE a function $\frac{f}{g}$, and gives $\frac{f'g-fg'}{g^{2}}$: it is a derivative computation. L'Hôpital's rule does not differentiate the quotient: it replaces the quotient $\frac{f}{g}$ with the NEW quotient $\frac{f'}{g'}$, where the numerator and the denominator have been differentiated SEPARATELY. On the example: $\lim_{x\to 0}\frac{\sin x}{x}$ has the form $\frac{0}{0}$, and we form $\frac{(\sin x)'}{(x)'}=\frac{\cos x}{1}\to 1$ ✓. With the quotient rule we would have obtained $\frac{x\cos x-\sin x}{x^{2}}$, an expression whose limit is much harder to evaluate and which, in any case, does not answer the question. The two rules have neither the same object nor the same purpose, and only their typographic resemblance brings them together.

The form is $0\cdot(-\infty)$. First rewriting, the bad one: $x\ln x=\frac{x}{1/\ln x}$, of the form $\frac{0}{0}$. Differentiating gives $\frac{1}{-\frac{1}{x(\ln x)^{2}}}=-x(\ln x)^{2}$, an expression WORSE than the starting one, since the power of the logarithm has gone up. Second rewriting, the good one: $x\ln x=\frac{\ln x}{1/x}$, of the form $\frac{-\infty}{+\infty}$. Differentiating: $\frac{1/x}{-1/x^{2}}=-x\to 0$. The limit is therefore $0$. The principle for choosing: put in the denominator the factor whose RECIPROCAL differentiates cleanly, that is, usually the power of $x$, and leave in the numerator the logarithm, which simplifies when differentiated. A bad choice does not give a wrong answer, it gives a computation that never ends.

Form $\infty\cdot 0$. The simplest is to substitute $t=\frac{1}{x}$, with $t\to 0^{+}$ as $x\to\infty$: the expression becomes $\frac{e^{t}-1}{t}$, of the form $\frac{0}{0}$. By L'Hôpital's rule: $\lim_{t\to 0}\frac{e^{t}}{1}=1$. The limit is $1$. You could also stay in $x$ and write $\frac{e^{1/x}-1}{1/x}$, then differentiate the numerator and the denominator: $\frac{e^{1/x}\cdot\left(-\frac{1}{x^{2}}\right)}{-\frac{1}{x^{2}}}=e^{1/x}\to e^{0}=1$ ✓, the factors $-\frac{1}{x^{2}}$ cancelling neatly. The result is predictable, by the way: $e^{t}-1\approx t$ near zero, which is the Taylor-expansion reading of the same equality.

Form $\infty-\infty$, so use a common denominator: $\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}$, of the form $\frac{0}{0}$. First round of L'Hôpital's rule: $\frac{\cos x-1}{\sin x+x\cos x}$, still $\frac{0}{0}$. Second round: $\frac{-\sin x}{\cos x+\cos x-x\sin x}=\frac{-\sin x}{2\cos x-x\sin x}$. Substitution now gives $\frac{0}{2}=0$, a determinate form: the limit is $0$. Interpretation: $\frac{1}{x}$ and $\frac{1}{\sin x}$ both blow up, but at the SAME speed to first order, and their difference stays finite. That is the deep meaning of the form $\infty-\infty$, where two infinities compete without any symbol telling which one wins.

Also a form $\infty-\infty$. L'Hôpital's rule is the wrong tool because there is no quotient, and manufacturing one here leads to heavier and heavier derivatives. The right tool is the CONJUGATE: $\sqrt{x^{2}+3x}-x=\frac{\left(\sqrt{x^{2}+3x}-x\right)\left(\sqrt{x^{2}+3x}+x\right)}{\sqrt{x^{2}+3x}+x}=\frac{3x}{\sqrt{x^{2}+3x}+x}$. Then factor $x$ out of the denominator: $\frac{3x}{x\left(\sqrt{1+\frac{3}{x}}+1\right)}=\frac{3}{\sqrt{1+\frac{3}{x}}+1}\to\frac{3}{2}$. The limit is $\frac{3}{2}$. The general lesson: L'Hôpital's rule is powerful but not always the shortest path, and for differences of square roots, the algebra of the conjugate remains unbeatable. A good reflex is to spend thirty seconds looking for an algebraic simplification before differentiating.

Near $a$, the linear approximation gives $f(x)\approx f(a)+f'(a)(x-a)=f'(a)(x-a)$, since $f(a)=0$; likewise $g(x)\approx g'(a)(x-a)$. The quotient becomes $\frac{f(x)}{g(x)}\approx\frac{f'(a)(x-a)}{g'(a)(x-a)}=\frac{f'(a)}{g'(a)}$, the factor $(x-a)$ cancelling BEFORE taking the limit. That is the whole idea of the rule: when two quantities vanish together, their ratio is decided by the SPEED at which each one vanishes, and that speed is exactly what the derivative measures. The common factor $(x-a)$, which is the source of the indeterminate form, is also what disappears when you simplify.

Cauchy's Mean Value Theorem: if $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in(a,b)$ such that $\left(f(b)-f(a)\right)g'(c)=\left(g(b)-g(a)\right)f'(c)$. Application to the $\frac{0}{0}$ case: assume $f(a)=g(a)=0$ and take $b=x$. The theorem gives $f(x)g'(c)=g(x)f'(c)$ for some $c$ between $a$ and $x$, which can be rewritten $\frac{f(x)}{g(x)}=\frac{f'(c)}{g'(c)}$ as soon as $g(x)$ and $g'(c)$ are nonzero. Now, as $x\to a$, the point $c$ is SQUEEZED between $a$ and $x$, so $c\to a$ as well. If $\frac{f'}{g'}$ has a limit $L$ at $a$, then $\frac{f'(c)}{g'(c)}\to L$, and therefore $\frac{f(x)}{g(x)}\to L$. The proof sheds light on two hypotheses that are often recited without being understood: $g'\neq 0$ is required so that we can divide, and $\lim\frac{f'}{g'}$ is required to EXIST because it is this limit, and it alone, that is carried over to $\frac{f}{g}$. If it does not exist, the chain breaks at its last link.

It is enough to set $t=\frac{1}{x}$: as $x\to+\infty$, $t\to 0^{+}$, and every limit at infinity becomes a limit at zero. On the example, $x=\frac{1}{t}$ gives $\frac{\ln x}{x}=\frac{\ln\left(\frac{1}{t}\right)}{\frac{1}{t}}=\frac{-\ln t}{\frac{1}{t}}=-t\ln t$, whose limit as $t\to 0^{+}$ is 0, as computed in exercise 3a. Numerical check: for $x=10^{6}$, $\frac{\ln x}{x}\approx 1.38\times 10^{-5}$, and for $t=10^{-6}$, $-t\ln t\approx 1.38\times 10^{-5}$: exactly the same value, as it should be since it is the same quantity written twice. The practical benefit is real: this substitution turns a $\frac{\infty}{\infty}$ form into a $0\cdot\infty$ form that is sometimes more convenient, and it also explains why the rule holds for $a=\pm\infty$ even though Cauchy's theorem only speaks of bounded intervals.

The proof is circular because applying L'Hôpital's rule requires differentiating $\sin x$, and therefore already knowing that $(\sin x)'=\cos x$. Now, the proof of that derivative goes through the definition $\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$, whose computation relies precisely on $\lim_{h\to 0}\frac{\sin h}{h}=1$. The result is thus used to prove itself: the argument is circular, and it proves nothing. The limit $\lim_{x\to 0}\frac{\sin x}{x}=1$ is established independently, by a geometric inequality on the unit circle, $\cos x\le\frac{\sin x}{x}\le 1$, followed by the Squeeze Theorem. It therefore comes BEFORE the whole theory of trigonometric derivatives. As for $\lim_{x\to 0}\frac{e^{x}-1}{x}$, there is nothing to compute: writing $\frac{e^{x}-1}{x}=\frac{e^{x}-e^{0}}{x-0}$, you recognize the difference quotient of the exponential function between 0 and $x$. Its limit is therefore, by the DEFINITION of the derivative, the number $\left(e^{x}\right)'$ evaluated at 0, that is $e^{0}=1$. The lesson holds for the whole chapter: many limits of the form $\frac{0}{0}$ are in fact derivatives in disguise, and recognizing them as such is instantaneous where L'Hôpital's rule requires a computation, when it does not go round in circles altogether.