Corrected exercises: L'Hôpital's rule and indeterminate forms (201-SN2-RE)

Form $\frac{0}{0}$. Differentiate: $\frac{4x^{3}}{3x^{2}}=\frac{4x}{3}\to\frac{8}{3}\approx 2.667$.

Form $\frac{0}{0}$. Differentiate: $\frac{\frac{2}{2\sqrt{2x+1}}}{1}=\frac{1}{\sqrt{2x+1}}\to\frac{1}{3}\approx 0.333$. Pitfall: forgetting the factor $2$ from the inner derivative, which gives $\frac{1}{6}$.

Form $\frac{0}{0}$. Differentiate: $\frac{2\sec^{2}(2x)}{5\cos(5x)}\to\frac{2}{5}$.

Form $\frac{1-1}{0}=\frac{0}{0}$. Differentiate: $\frac{3e^{3x}+e^{-x}}{2\cos 2x}\to\frac{3+1}{2}=2$. Pitfall: $\left(e^{-x}\right)'=-e^{-x}$, and the minus sign of the difference turns it into a plus.

Form $\frac{0}{0}$, since $\ln 1=0$. Differentiate: $\frac{\frac{1}{x}}{2x}=\frac{1}{2x^{2}}\to\frac{1}{2}$.

Form $\frac{0}{0}$. Differentiate: $\frac{2x}{3\sin 3x}$, still $\frac{0}{0}$; differentiate again: $\frac{2}{9\cos 3x}\to\frac{2}{9}\approx 0.222$. Pitfall: stopping after the first round, or forgetting the $3$ at each differentiation of the cosine.

Form $\frac{\infty}{\infty}$. Differentiate: $\frac{3e^{3x}}{2x}$, still $\frac{\infty}{\infty}$; differentiate again: $\frac{9e^{3x}}{2}\to+\infty$. The exponential beats any power.

Form $\frac{\infty}{\infty}$. Differentiate: $\frac{\frac{1}{x}}{\frac{1}{3}x^{-\frac{2}{3}}}=\frac{3}{x^{\frac{1}{3}}}\to 0$. The logarithm loses against any positive power, even $x^{\frac{1}{3}}$.

Form $0\cdot\infty$. Turn the tangent into a cotangent: $\frac{1-x}{\cot\left(\frac{\pi x}{2}\right)}$, of the form $\frac{0}{0}$. Differentiate: $\frac{-1}{-\frac{\pi}{2}\csc^{2}\left(\frac{\pi x}{2}\right)}\to\frac{2}{\pi}\approx 0.637$. Pitfall: the factor $\frac{\pi}{2}$ from the inner derivative.

Form $\infty-\infty$. Common denominator: $\frac{x-e^{x}+1}{x\left(e^{x}-1\right)}$, of the form $\frac{0}{0}$. Differentiate: $\frac{1-e^{x}}{e^{x}-1+xe^{x}}$, still $\frac{0}{0}$; differentiate again: $\frac{-e^{x}}{2e^{x}+xe^{x}}\to-\frac{1}{2}$.

Form $1^{\infty}$. Set $y=(\cos x)^{\frac{1}{x^{2}}}$, so $\ln y=\frac{\ln(\cos x)}{x^{2}}$, of the form $\frac{0}{0}$. Differentiate: $\frac{-\tan x}{2x}$, still $\frac{0}{0}$; differentiate again: $\frac{-\sec^{2}x}{2}\to-\frac{1}{2}$. Therefore $y\to e^{-\frac{1}{2}}\approx 0.607$. Pitfall: answering $-\frac{1}{2}$, the limit of the logarithm and not that of $y$.

Form $0^{0}$, since $\ln x\to-\infty$ makes the exponent tend to zero. Set $\ln y=\frac{3\ln x}{1+\ln x}$, of the form $\frac{\infty}{\infty}$. Differentiate: $\frac{\frac{3}{x}}{\frac{1}{x}}=3$. Therefore $y\to e^{3}\approx 20.086$. Pitfall: believing that a $0^{0}$ form is always worth $1$.

You recognize $\dfrac{f(x)-f(0)}{x-0}$ with $f(x)=\sqrt{1+x}$ and $f(0)=1$. The limit is therefore $f'(0)$. Now $f'(x)=\dfrac{1}{2\sqrt{1+x}}$, so $f'(0)=\dfrac{1}{2}$. No manipulation, no differentiation of the quotient: the table of derivatives has already done the work.

First limit: $\dfrac{x^{5}-2^{5}}{x-2}$ is the difference quotient of $f(x)=x^{5}$ between $2$ and $x$, so the limit is $f'(2)=5\times 2^{4}=80$. Second: $\dfrac{\sin\left(\frac{\pi}{6}+h\right)-\sin\frac{\pi}{6}}{h}$ is the difference quotient of the sine at $\dfrac{\pi}{6}$, so the limit is $\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\approx 0.866$. The signal to remember: a $-$ followed by the value of the function at the point, divided by the distance to the point, almost always announces a derivative.

$\dfrac{x+1}{x^{2}+3}$ at $0$ equals $\dfrac{1}{3}$ by substitution: DETERMINATE. $\dfrac{\ln x}{x}$ at $0^{+}$ gives $\dfrac{-\infty}{0^{+}}$, which is not an indeterminate form: the limit is $-\infty$. The other two are indeterminate: $\dfrac{\infty}{\infty}$ for the third, $\dfrac{0}{0}$ for the fourth. Watch the trap in the second one: $\dfrac{0}{0}$ is indeterminate, but $\dfrac{\text{nonzero}}{0}$ never is.

$\lim_{x\to\infty}\dfrac{2x^{3}+x}{7x^{3}-4}$ is settled by ALGEBRA, factoring $x^{3}$ out of the top and the bottom: the limit is $\dfrac{2}{7}$. L'Hôpital's rule would give the same result in three differentiations, which would be a waste. $\lim_{x\to 0}\dfrac{1-\cos x}{x}$ is indeed $\dfrac{0}{0}$; differentiating top and bottom, $\dfrac{\sin x}{1}\to 0$. The limit is $0$, and the example shows that an indeterminate form can perfectly well have a finite limit, zero included.

The sentence confuses the FORM and the LIMIT. “Indeterminate” describes the expression obtained by substitution, not the behaviour of the function: it only means that substitution is not enough to conclude, and that one more tool is needed. The three limits $\lim_{x\to 0}\dfrac{x^{2}}{x}=0$, $\lim_{x\to 0}\dfrac{x}{x}=1$ and $\lim_{x\to 0}\dfrac{x}{x^{3}}=\infty$ are all of the form $\dfrac{0}{0}$ and have three different behaviours: that is exactly what the word “indeterminate” describes. Correct wording: “substitution gives the indeterminate form $\dfrac{0}{0}$, so the indeterminate form must be resolved”, then you state the result. A limit can also fail to exist, like $\lim_{x\to 0}\sin\dfrac{1}{x}$, but that is not the same thing, and it has to be proved.

At $x=0$: the numerator equals $1-1-0=0$ and the denominator $0$: form $\dfrac{0}{0}$, the rule applies. Differentiate: $\dfrac{2e^{2x}-2}{2x}$, which is still $\dfrac{0}{0}$ at $0$: CHECK, and start over. $\dfrac{4e^{2x}}{2}\to\dfrac{4}{2}=2$. The limit is $2$. The move you must never skip is the intermediate check: you may apply the rule again as many times as the indeterminate form PERSISTS, but only once you have made sure of it.

The mistake is to apply the rule without checking for an indeterminate form. At $x=1$, the numerator equals $4$ and the denominator $2$: the form is $\dfrac{4}{2}$, perfectly determinate, and since the function is continuous at $1$, the limit is obtained by substitution: it is $2$. The student also finds $2$, but through wrong reasoning: it is a coincidence, and the same procedure on $\lim_{x\to 1}\dfrac{x^{2}+3}{x+3}$ would give $\dfrac{2}{1}=2$ instead of $1$. A method that lands on the right answer by chance is worth zero marks.

The quotient rule is used to DIFFERENTIATE a function that is a quotient; L'Hôpital's rule is used to compute a LIMIT of a quotient, and it requires differentiating the top and the bottom SEPARATELY. The two have neither the same purpose nor the same result: here the quotient rule would give $\dfrac{xe^{x}-(e^{x}-1)}{x^{2}}$, which is the derivative of the function, not its limit at $0$. The confusion comes from the fact that both statements talk about “differentiating a quotient”; the cure is to ask yourself each time what you are looking for, a derivative function or a limit value.

Because the derivative of the sine was ESTABLISHED from the limit $\lim_{x\to 0}\dfrac{\sin x}{x}=1$, proved by the Squeeze Theorem in the chapter on limits. Using $(\sin u)'=\cos u$ to prove again a limit of this family assumes what you want to prove. The result, $3$, is indeed correct; it is the PROOF that is flawed. The proper write-up sets $u=3x$ and invokes the special limit: $\dfrac{\sin 3x}{x}=3\cdot\dfrac{\sin 3x}{3x}\to 3\times 1=3$. The same warning applies to $\lim_{x\to 0}\dfrac{e^{x}-1}{x}$ and $\lim_{x\to 0}\dfrac{\ln(1+x)}{x}$, which ARE the derivatives that were used to build the table.

The rule of conduct comes in two steps. First: CHECK the indeterminate form before every differentiation, never just once at the start. Second: before applying the rule, ask yourself whether the limit is not itself the derivative that was used to establish the derivative you are about to use; if so, conclude directly from the definition or from the special limit, without going through L'Hôpital's rule.

Form $0\cdot(-\infty)$. First rewriting, $\dfrac{\sqrt{x}}{1/\ln x}$: the derivative of the denominator is $-\dfrac{1}{x(\ln x)^{2}}$, and the quotient of derivatives becomes $-\dfrac{x(\ln x)^{2}}{2\sqrt{x}}$, worse than at the start. Second rewriting, $\dfrac{\ln x}{x^{-1/2}}$, of the form $\dfrac{-\infty}{+\infty}$: differentiate, $\dfrac{1/x}{-\frac{1}{2}x^{-3/2}}=-2x^{1/2}\to 0$. The limit is $0$. The logarithm SIMPLIFIES when differentiated, the root does not: the logarithm is the one to keep on top.

Form $\infty\cdot 0$, since $\arctan x\to\dfrac{\pi}{2}$. Move the factor $x$ down: $\dfrac{\frac{\pi}{2}-\arctan x}{1/x}$, of the form $\dfrac{0}{0}$. Differentiate: $\dfrac{-\frac{1}{1+x^{2}}}{-\frac{1}{x^{2}}}=\dfrac{x^{2}}{1+x^{2}}\to 1$. The limit is $1$, which can also be read this way: the gap between $\arctan x$ and its asymptote decreases like $\dfrac{1}{x}$.

Form $\infty-\infty$, both terms diverging at $1$. Use a common denominator: $\dfrac{(x-1)-\ln x}{(x-1)\ln x}$, of the form $\dfrac{0}{0}$. Differentiate: $\dfrac{1-\frac{1}{x}}{\ln x+\frac{x-1}{x}}=\dfrac{x-1}{x\ln x+x-1}$, still $\dfrac{0}{0}$. Differentiate again: $\dfrac{1}{\ln x+1+1}\to\dfrac{1}{2}$. The limit is $\dfrac{1}{2}$. Going to a common denominator is compulsory: differentiating each term separately makes no sense, the rule is about a QUOTIENT.

Form $\infty-\infty$. The conjugate: $x-\sqrt{x^{2}-4x}=\dfrac{x^{2}-(x^{2}-4x)}{x+\sqrt{x^{2}-4x}}=\dfrac{4x}{x+\sqrt{x^{2}-4x}}$. Dividing top and bottom by $x$, the denominator tends to $2$ and the limit is $2$. L'Hôpital's rule would be the wrong tool for two reasons: you would first have to rewrite the difference as a quotient, which is already the conjugate's job; and differentiating a square root gives back a square root, so that the expression never simplifies and the rule goes in circles. Practical rule: faced with a difference of two roots, or of a root and a polynomial, ALWAYS try the conjugate first.

The criterion: move to the denominator the factor whose derivative COMPLICATES the expression, and keep on top the one whose derivative simplifies it. A logarithm simplifies when differentiated, and so does a power; an exponential and a root do not. When in doubt, try, and give up as soon as the quotient of derivatives is heavier than the original.

Form $\dfrac{0}{0}$. Differentiate: $\dfrac{\frac{1}{1+x^{2}}-1}{3x^{2}}=\dfrac{-x^{2}}{3x^{2}(1+x^{2})}=\dfrac{-1}{3(1+x^{2})}\to-\dfrac{1}{3}$. Simplifying ALGEBRAICALLY after the first differentiation saves two rounds of the rule: it always pays to check whether the quotient of derivatives simplifies before differentiating again.

Form $\dfrac{0}{0}$. Differentiate: $\dfrac{\frac{1}{1+x}-1}{2x}=\dfrac{-x}{2x(1+x)}=\dfrac{-1}{2(1+x)}\to-\dfrac{1}{2}$. Same remark: the simplification falls into place by itself.

Form $\infty\cdot 0$. Set $u=\dfrac{1}{x}$, which tends to $0^{+}$: the expression becomes $\dfrac{e^{u}-1-u}{u^{2}}$, of the form $\dfrac{0}{0}$. Differentiate: $\dfrac{e^{u}-1}{2u}$, still $\dfrac{0}{0}$; differentiate again: $\dfrac{e^{u}}{2}\to\dfrac{1}{2}$. The change of variable is safer here than moving a factor down directly, because it replaces a limit at infinity by a limit at zero, where forms are easier to read.

Form $\dfrac{\infty}{\infty}$. Differentiate: $\dfrac{\frac{2x}{x^{2}+1}}{\frac{3x^{2}+1}{x^{3}+x}}=\dfrac{2x(x^{3}+x)}{(x^{2}+1)(3x^{2}+1)}=\dfrac{2x^{2}(x^{2}+1)}{(x^{2}+1)(3x^{2}+1)}=\dfrac{2x^{2}}{3x^{2}+1}\to\dfrac{2}{3}$. You could also conclude without differentiating: $\ln(x^{2}+1)\sim 2\ln x$ and $\ln(x^{3}+x)\sim 3\ln x$, so the quotient tends to $\dfrac{2}{3}$. Both routes are accepted; the second is faster and more robust.

Each limit compares a function with its linear approximation at $0$, that is, with its TANGENT LINE. For the arctangent, $\arctan x-x\approx-\dfrac{x^{3}}{3}$: the gap is negative for $x>0$ and positive for $x<0$, so the curve crosses its tangent line, the sign of an INFLECTION point at $0$. For the logarithm, $\ln(1+x)-x\approx-\dfrac{x^{2}}{2}$: the gap is negative on BOTH sides, so the curve stays below its tangent line, which corresponds to a graph that is concave down. For the exponential, $e^{u}-1-u\approx+\dfrac{u^{2}}{2}$: the curve stays ABOVE its tangent line, concave up. An indeterminate limit of this kind therefore gives more than a number: it tells you on which side of its tangent line the curve lies, and how fast it moves away from it. That is exactly the information you will use in the chapter on curve sketching.

Form $0^{0}$, since $\tan x\to 0^{+}$ and $x\to 0^{+}$. Set $y=(\tan x)^{x}$, so $\ln y=x\ln(\tan x)$, of the form $0\cdot(-\infty)$. Move a factor down: $\dfrac{\ln(\tan x)}{1/x}$, of the form $\dfrac{-\infty}{+\infty}$. Differentiate: $\dfrac{\frac{1+\tan^{2}x}{\tan x}}{-\frac{1}{x^{2}}}=-\dfrac{x^{2}(1+\tan^{2}x)}{\tan x}$. Since $\tan x\sim x$ near $0$, this behaves like $-x\to 0$. Therefore $\ln y\to 0$ and $y\to e^{0}=1$.

Form $1^{\infty}$, the base tending to $1$ and the exponent to infinity. Set $\ln y=x\ln\dfrac{x+2}{x-1}$, of the form $\infty\cdot 0$. Move a factor down: $\dfrac{\ln(x+2)-\ln(x-1)}{1/x}$, of the form $\dfrac{0}{0}$. Differentiate: $\dfrac{\frac{1}{x+2}-\frac{1}{x-1}}{-\frac{1}{x^{2}}}=\dfrac{\frac{-3}{(x+2)(x-1)}}{-\frac{1}{x^{2}}}=\dfrac{3x^{2}}{(x+2)(x-1)}\to 3$. Therefore $\ln y\to 3$ and $y\to e^{3}\approx 20.09$.

Form $\infty^{0}$, since $\dfrac{1}{x}\to+\infty$ and $\sin x\to 0$. Set $\ln y=\sin x\cdot\ln\dfrac{1}{x}=-\sin x\ln x$. Since $\sin x\sim x$, this behaves like $-x\ln x$, whose limit is $0$. Therefore $y\to e^{0}=1$. You could also write $\dfrac{-\ln x}{1/\sin x}$ and differentiate, but recognizing $\sin x\sim x$ shortens everything.

The reasoning confuses a base EQUAL to $1$ with a base that TENDS to $1$. If the base were exactly $1$, the power would be $1$ whatever the exponent. But here the base is $1+\dfrac{3}{x-1}$, so it exceeds $1$ by a little, and that excess is raised to the power $x$, which tends to infinity. Two opposite tendencies fight it out, and the outcome depends on their RATES: that is exactly what the word “indeterminate” describes. The computation settles it: the limit is $e^{3}\approx 20.09$, twenty times the announced answer. An immediate counterexample to convince yourself: $\left(1+\dfrac{1}{x}\right)^{x}\to e$, not $1$.

The right answer is $1$, and the skipped step is the last one: the method computes the limit of the LOGARITHM of the expression, not that of the expression. You must therefore go back to the exponential, $y=e^{\ln y}\to e^{0}=1$. The mistake is all the easier to make because the auxiliary variable $y$ often vanishes from the paper; the cure is to write it explicitly on the first line, “let $y=\ldots$”, and to finish with “therefore $y\to e^{L}$”. A marker takes off one or two marks for this omission, which turns $e^{3}\approx 20$ into $3$ in case b).

Form $\dfrac{\infty}{\infty}$. Differentiate: $\dfrac{2\ln x\cdot\frac{1}{x}}{1}=\dfrac{2\ln x}{x}$, still $\dfrac{\infty}{\infty}$. Differentiate again: $\dfrac{2/x}{1}=\dfrac{2}{x}\to 0$. The limit is $0$. The mechanism is clear: each differentiation lowers the power of the logarithm by one notch without touching the $x$ at the bottom, so the logarithm always runs out in the end. The same reasoning gives $\lim\dfrac{(\ln x)^{n}}{x^{p}}=0$ for all $n$ and $p>0$.