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The rule, the two forms it applies to, the five that must be rearranged first, and the ones it must never touch.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to name the form of a limit before doing anything else, apply L'Hopital's rule to a $0/0$ or $\infty/\infty$ quotient by differentiating numerator and denominator separately, re-check the form after each application and stop when it is determinate, rearrange a product, a difference or a power into a quotient the rule can reach — exponentiating at the end for a power form — recognise the determinate forms the rule must not be applied to, and notice when substitution or algebra would have been quicker.
L'Hopital's rule. If $\lim \dfrac{f}{g}$ is of the form $\dfrac00$ or $\dfrac{\pm\infty}{\pm\infty}$, and $\lim \dfrac{f'}{g'}$ exists (or is $\pm\infty$), then $$\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}.$$
It holds as $x \to a$, $x \to a^{\pm}$ and $x \to \pm\infty$ alike.
It is not the quotient rule. Differentiate the numerator and the denominator separately — do not differentiate the quotient. The two produce different expressions and only one of them is this rule.
The seven indeterminate forms.
| Form | What to do |
|---|---|
| $\dfrac00$, $\dfrac{\infty}{\infty}$ | apply the rule directly |
| $0 \cdot \infty$ | move a factor into the denominator to make a quotient |
| $\infty - \infty$ | combine over a common denominator, or rationalise |
| $0^0$, $1^\infty$, $\infty^0$ | take logarithms, find the limit of the log, exponentiate |
Only the first row is within the rule's reach as written. Everything else must be rearranged into a quotient first.
*What is not indeterminate*, and where the rule therefore says nothing: $\dfrac{c}{0}$ with $c \ne 0$ (infinite), $\dfrac{0}{c}$ (zero), $0^\infty$ (zero), $\infty^\infty$ (infinite). Applied to one of these the rule produces a number, and the number is usually wrong.
Another way: picture
Two functions both heading to zero at $a$. Which one gets there faster decides the limit of their ratio, and "how fast" is what a derivative measures — so comparing the derivatives compares the rates of approach. That is the whole intuition, and the honest version is the Cauchy mean value theorem, from which the rule is proved.
Another way: steps
L'Hopital is powerful and frequently unnecessary, and reaching for it first can turn a one-line limit into a long one.
$$\lim_{x\to\infty}\frac{3x^2 + 1}{5x^2 - x}$$ is $\infty/\infty$, and the rule applies — twice, giving $\dfrac{6}{10}$. Dividing by $x^2$ gives the same answer in one step and shows why it is the ratio of the leading coefficients.
$$\lim_{x\to 3}\frac{x^2-9}{x-3}$$ is $0/0$, and the rule gives $\dfrac{2x}{1} \to 6$. Factoring gives it just as fast and without a theorem.
And sometimes the rule simply fails to help: $$\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$$ differentiates to $\dfrac{\sqrt{x^2+1}}{x}$, which is the reciprocal of what you started with. Applying it again returns the original. The loop is unbreakable, and dividing by $x$ settles it immediately.
So: try substitution, then algebra, and reach for L'Hopital when those have failed.
Applying it to a form that is not indeterminate. The rule's one hypothesis, and the one most often skipped. $\lim_{x \to 0}\dfrac{\cos x}{x+1}$ is $\dfrac11 = 1$; the rule would give $\dfrac{-\sin x}{1} \to 0$, which is wrong and looks like working.
Using the quotient rule by accident. Differentiate the two separately.
Forgetting to re-check the form. Each application needs its own licence, and one application too many differentiates a determinate quotient into nonsense.
Applying it to a product or a difference. Rearrange into a quotient first; there is no product version.
Reporting the logarithm's limit. For a power form, $\ln L$ is what the rule found and $L = e^{\ln L}$ is the answer.
Concluding the limit does not exist when $\lim f'/g'$ does not. The rule is one-directional. If the derivative ratio has no limit, the rule is silent — the original limit may still exist, and $\lim_{x\to\infty}\dfrac{x + \sin x}{x}$ is $1$ while $\dfrac{1 + \cos x}{1}$ has no limit at all.
Most mistakes in calculus announce themselves — a nonsensical value, an impossible sign, an expression that will not simplify. Applying L'Hopital's rule to a determinate form does not. It differentiates two perfectly good functions, divides them, and hands you a number, with working that looks exactly like working that is right.
$\lim_{x\to 0}\dfrac{\cos x}{x + 1}$: substitution gives $1$. The rule gives $\dfrac{-\sin x}{1} \to 0$. Both calculations are short, and only one of them was allowed.
So the habit is not optional and it is one line: substitute, and name the form, before writing anything else. If the form is not $0/0$ or $\infty/\infty$, the rule is not available — and the substitution you just did has very often given you the answer already.
$\lim_{x \to 0}\left(\dfrac{1}{x} - \dfrac{1}{\sin x}\right)$ is $\infty - \infty$: indeterminate, and not a quotient.
Rearrange before applying anything.
Common denominator: $\dfrac{\sin x - x}{x\sin x}$, now $0/0$. Rule: $\dfrac{\cos x - 1}{\sin x + x\cos x}$ — still $0/0$.
Check the form again.
Again: $\dfrac{-\sin x}{2\cos x - x\sin x}$, which at $0$ is $\dfrac{0}{2} = 0$. Determinate, so stop. The limit is $0$.
Three checks, two applications.
$\lim_{x\to\infty}\left(1 + \dfrac{1}{x}\right)^x$ is $1^\infty$ — indeterminate, despite $1$ to any power being $1$, because the base is only approaching $1$.
Power form: take logarithms.
$\ln L = \lim x\ln\left(1 + \dfrac1x\right)$, a $\infty \cdot 0$ product. As a quotient: $\dfrac{\ln(1 + 1/x)}{1/x}$, of form $0/0$.
Two rearrangements before the rule.
The rule gives $\dfrac{-1/x^2 \cdot \frac{1}{1+1/x}}{-1/x^2} = \dfrac{1}{1 + 1/x} \to 1$. So $\ln L = 1$ and $L = e$ — the definition of $e$, recovered.
Exponentiate; the answer is $e$, not $1$.
Substituting: $\dfrac{1 - 1 - 0}{0} = \dfrac00$. Indeterminate, so the rule applies.
Check first.
Once: $\dfrac{e^x - 1}{2x}$, still $0/0$. Again: $\dfrac{e^x}{2}$.
Re-check after each.
At $0$: $\dfrac12$. Determinate now, so stop.
Match each limit form to the correct response.
| Differentiate numerator and denominator if hypotheses hold | Differentiate numerator and denominator if hypotheses hold | Not an indeterminate form; analyze divergence or one-sided signs | Rewrite as a quotient before considering the rule | |
|---|---|---|---|---|
| $0/0$ | ||||
| $\infty/\infty$ | ||||
| $5/0$ | ||||
| $0\cdot\infty$ |
Match each limiting form or expression to the correct L'Hopital action.
| Differentiate numerator and denominator separately if hypotheses hold | The same rule may apply after checking its hypotheses | Rewrite as $\ln x/(1/x)$ before applying the rule | Take logarithms, find the log limit, then exponentiate | |
|---|---|---|---|---|
| A quotient tends to $0/0$ | ||||
| A quotient tends to $\infty/\infty$ | ||||
| $x\ln x$ as $x\to0^+$ | ||||
| $f(x)^{g(x)}$ with form $1^\infty$ |
Find $\displaystyle\lim_{x \to 0} \frac{7x + x^2}{4x - x^3}$.
Answer:
Evaluate $\lim_{x \to 0} \dfrac{\ln(1 + x)}{x}$, to four decimal places where it is not a whole number.
Answer:
Should L'Hopital's rule be used on $\lim_{x \to 0} \dfrac{\cos x}{x + 1}$?
Find $\displaystyle\lim_{x \to 0^+} x^x$.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find $\displaystyle\lim_{x \to 0} \frac{5x^2}{2\bigl(1 - \cos(2x)\bigr)}$.
Answer:
You can apply the rule where it is permitted and say where it is not. Say in your own words why applying it to a determinate form is more dangerous than most errors. Next: antiderivatives, and the start of the integral.
8. Your turn: $\lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2}$, step 3