L'Hôpital's Rule

${\displaystyle \frac{0}{0}}$ — ${\displaystyle \underset{x\to 0}{lim}\frac{e^x-1}{x}=1}$. The template case. Differentiating gives $e^x/1\to 1$, and the values $1.0517$, $1.00502$, $1.00050$ confirm it. No algebra clears this one: $e^x-1$ has no factor of $x$ to cancel.

${\displaystyle \frac{0}{0}}$ with a rate ratio — ${\displaystyle \underset{x\to 0}{lim}\frac{\sin 3x}{x}=3}$. Differentiating gives $3\cos 3x\to 3$. The answer is the ratio of the speeds at which the two functions leave zero, and the 3 comes from the chain rule on the numerator.

${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$ — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{\ln x}{x}=0}$. Differentiating gives $\frac{1/x}{1}\to 0$. The values fall $0.2303$, $0.04605$, $0.000921$, $0.0000138$ at $x=10,{10}^2,{10}^4,{10}^6$. This is the standard statement that logarithmic growth loses to linear growth, and the same argument beats any positive power.

${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$ needing two passes — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}=0}$. One application gives $\frac{2x}{e^x}$, still indeterminate; a second gives $\frac{2}{e^x}$, which is not, so the limit is read directly. The values $0.1684$, $0.00454$, $8.24\times {10}^{-7}$ at $x=5,10,20$ show the exponential winning comfortably. For $x^n/e^x$ the same argument takes $n$ passes.

$0\cdot \mathrm{\infty }$ — ${\displaystyle \underset{x\to 0^{+}}{lim}x\ln x=0}$. Not a quotient, so it is rewritten as $\frac{\ln x}{1/x}$ before the rule applies. The values $-0.2303$, $-0.04605$, $-0.0069$, $-0.0000138$ climb to 0 from below: $x$ reaches zero faster than $\ln x$ reaches $-\mathrm{\infty }$.

**A quotient that is not indeterminate, ${\displaystyle \underset{x\to 0}{lim}\frac{x+1}{x+2}=\frac{1}{2}}$.** Substitution settles it, and the rule must not be used. Applying it anyway returns $1/1=1$, wrong by a factor of two. It sits beside the others as the case that looks identical in shape and differs entirely in what it permits.

Where the rule is silent — ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x+\sin x}{x}=1}$. The form is $\mathrm{\infty }/\mathrm{\infty }$, and the remaining conditions hold: both parts are differentiable and the denominator's derivative is the constant $1$. So the rule applies legitimately, and differentiating gives $1+\cos x$, which oscillates between 0 and 2 and has no limit. The theorem therefore says nothing. The original limit exists regardless, since $\frac{\sin x}{x}\to 0$ leaves $1+0$.

Two forms are handled directly, one after a rewrite, one only by stopping at the right moment, and two not at all, once because the hypothesis fails and once because the conclusion's condition fails. Reading which case is in front of you is the skill; the differentiation is the easy part.

Applying it to a determinate quotient. ${\displaystyle \underset{x\to 0}{lim}\frac{x+1}{x+2}}$. Substitution gives $\frac{1}{2}$. The numerator tends to 1, not 0, so the form is not indeterminate and the rule's hypothesis fails.

Applying it anyway gives $\frac{1}{1}=1$. The values $0.5238$, $0.5025$, $0.5000$ at $x=0.1,0.01,0$ show which answer is right. Nothing about the functions is pathological: both are differentiable everywhere and the denominator's derivative never vanishes. The single missing hypothesis is enough to make the conclusion false.

This is the error to guard against, because it produces a plausible wrong number rather than an obvious failure.

Using the quotient rule by mistake. For $\underset{x\to 0}{lim}\frac{e^x-1}{x}$, the rule calls for $\frac{(e^x-1{)}^{\prime }}{(x{)}^{\prime }}=\frac{e^x}{1}$. Computing instead the derivative of the quotient,

answers a different question entirely. It is the slope of the function whose limit was wanted, not the limit. The theorem replaces a quotient by a quotient of derivatives; it does not differentiate the quotient.

Continuing after the form becomes determinate. For $\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}$, two applications give $\frac{2}{e^x}$, whose limit is 0. A third application would give $\frac{0}{e^x}=0$. The same answer here by luck, since the numerator is constant. But the habit is the Step 1 error in disguise, and on a quotient like $\frac{x+1}{x+2}$ reached mid-problem it changes the answer.

The discipline is to re-check the form before every application, not only the first.

Concluding non-existence when the rule is silent. ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{x+\sin x}{x}}$. The form is $\mathrm{\infty }/\mathrm{\infty }$ and the remaining conditions hold, so the rule applies, and differentiating gives $1+\cos x$, which oscillates between 0 and 2 forever. No limit.

It does not follow that the original limit fails to exist. Writing the expression as $1+\frac{\sin x}{x}$ and using $|\sin x|\le 1$ gives $\frac{\sin x}{x}\to 0$ by squeezing, so the limit is exactly 1. The theorem's third hypothesis is a condition on its conclusion: when $f^{\prime }/g^{\prime }$ has no limit, the theorem is simply inapplicable, and says nothing in either direction.

Using it circularly. $\underset{x\to 0}{lim}\frac{\sin x}{x}$ is $0/0$, and the rule gives $\frac{\cos x}{1}\to 1$, which is correct. But $\frac{d}{dx}\sin x=\cos x$ is proved from this limit, so the argument assumes what it sets out to establish. The correct route is the squeeze theorem with $\cos x\le \frac{\sin x}{x}\le 1$, whose bounds both tend to 1, checked at $x=0.1$: $0.99500\le 0.99833\le 1$.

The rule is not wrong here; it is unavailable as a proof, and that distinction decides what the calculation establishes.

L'Hôpital's rule applies to any quotient of differentiable functions, so a limit of a fraction can always be found by differentiating the top and bottom.