A rule you have to earn the right to use AB & BC

L'Hospital's Rule. If $\displaystyle\lim_{x \to c}\frac{f(x)}{g(x)}$ has the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and the pieces are differentiable near $c$ with $g'(x) \neq 0$, then

$$\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)},$$

provided that second limit exists.

The first clause is a requirement, not a description. Substitute first and read what you get:

1. $\frac{0}{0}$ or $\frac{\infty}{\infty}$: the rule applies.
2. $\frac{5}{7}$ or any ratio of finite nonzero numbers: the limit is that number. There is nothing to resolve.
3. $\frac{5}{0}$: the limit is infinite or one-sided, and the rule does not apply.
4. $\frac{0}{5}$: the limit is 0, and the rule does not apply.

Applied to a determinate form the rule returns a number, confidently and wrongly. For $\displaystyle\lim_{x\to 2}\frac{x^{2}-4}{x-1}$ substitution gives $\frac{0}{1} = 0$, which is the answer. Differentiating top and bottom gives $\frac{2x}{1} \to 4$, which is not.

The rule differentiates the numerator and the denominator as two independent functions. It is not the quotient rule and it is not related to it:

$$\frac{f'}{g'} \qquad \text{not} \qquad \frac{f'g - fg'}{g^{2}}.$$

The resemblance is entirely superficial and it costs points every year. For $\displaystyle\lim_{x\to 0}\frac{e^{x}-1}{x}$, the rule gives $\frac{e^{x}}{1} \to 1$. The quotient rule would give $\frac{xe^{x} - (e^{x}-1)}{x^{2}}$, which is the derivative of a completely different object and is still $\frac{0}{0}$ at $x = 0$.

Two habits that keep this straight. Write $f$ and $g$ on separate lines before differentiating anything. And remember what the rule is doing: near $c$ both functions are approximately their own linearizations, so the ratio is approximately $\frac{f'(c)(x-c)}{g'(c)(x-c)}$, and the steps cancel. That is Topic 4.6 explaining Topic 4.7.

One application often leaves the same indeterminate form, and the rule can be applied again as long as the form still qualifies. Recheck between applications rather than assuming.

$$\lim_{x\to 0}\frac{1-\cos x}{x^{2}} \;\overset{0/0}{=}\; \lim_{x\to 0}\frac{\sin x}{2x} \;\overset{0/0}{=}\; \lim_{x\to 0}\frac{\cos x}{2} = \frac{1}{2}.$$

Stopping after one step and substituting into $\frac{\sin x}{2x}$ gives $\frac{0}{0}$, which is not an answer, or gives 0 if the $\frac{0}{0}$ is misread as zero. A numerical check settles it: at $x = 0.01$ the original expression is $0.4999958$.

The other indeterminate forms have to be rewritten as a quotient before the rule is available at all:

1. $0 \cdot \infty$: move one factor into the denominator. $\displaystyle\lim_{x\to 0^{+}} x\ln x = \lim_{x\to 0^{+}}\frac{\ln x}{1/x}$, now $\frac{-\infty}{\infty}$, which gives $\frac{1/x}{-1/x^{2}} = -x \to 0$.
2. $\infty - \infty$: combine over a common denominator first.
3. $1^{\infty}$, $0^{0}$, $\infty^{0}$: take a logarithm, resolve the resulting limit, then exponentiate.

What never works is applying the rule to a product as though it were a quotient.

When the rule delivers a number, that number is the limit of the original quotient. It is not a statement about derivatives, and it is not a value of the original function.

From $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1} = 1$, the conclusion is that the values of $\frac{\sin x}{x}$ get arbitrarily close to 1 as $x$ approaches 0. Three things it does not say:

1. That $\frac{\sin x}{x}$ equals 1 at $x = 0$. It is undefined there, which is why a limit was needed.
2. That the derivative of $\frac{\sin x}{x}$ approaches 1. The $\frac{\cos x}{1}$ was a computational detour, not the object being described.
3. Anything about $\sin x$ or $x$ separately. Both approach 0; only their ratio approaches 1.

The same care applies in reverse. If $f(3) = 0$ and $f'(3) = -2$, then $\displaystyle\lim_{x\to 3}\frac{f(x)}{x-3} = f'(3) = -2$, which is the definition of the derivative from Topic 2.2 wearing a different hat. The $-2$ is a rate, and it does not say that $f(3) = -2$.