Using L'Hospital's Rule for Determining Limits of Indeterminate Forms
Introduced| English |
|---|
| L'Hospital's Rule/ˈelhɒspɪtlz ruːl/ |
| indeterminate forms/ˌɪndɪˈtɜːmɪnət fɔːmz/ |
A derivative trick for stubborn limits
- Back in Unit 1, $\tfrac00$ meant "simplify by algebra." Sometimes no algebra helps.
- L'Hospital's Rule 洛必达法则 uses derivatives to crack those stubborn limits.
- It applies to the indeterminate forms $\tfrac00$ and $\tfrac{\infty}{\infty}$.
- The idea: replace the ratio of functions by the ratio of their derivatives.
Check the form first
- L'Hospital only applies to a genuine indeterminate form: $\dfrac00$ or $\dfrac{\infty}{\infty}$.
- Substitute first to confirm you have one — if direct substitution already gives a number, you are done, and the rule does not apply.
- Using it on a non-indeterminate limit (like $\tfrac{5}{2}$) gives a wrong answer.
- So always verify the form before reaching for the rule.
L'Hospital's Rule applies to which indeterminate forms?
Only $\tfrac00$ and $\tfrac\infty\infty$. $\tfrac52$ is already determinate; $\tfrac30$ is an asymptote, not indeterminate.
Before applying the rule, confirm the limit is an ____ form by substituting.
Only $\tfrac00$ or $\tfrac\infty\infty$ qualify.
Differentiate top and bottom separately
- The rule: $$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$(when the form is indeterminate).
- Warning: this is not the quotient rule — you differentiate $f$ and $g$ independently, no minus, no $g^2$.
- Then re-evaluate the new, simpler limit.
- $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=1$ — the famous limit, in one line.
A 0/0 ratio with a real limit
y = sin(x)/x (near 0)
$\tfrac{\sin x}{x}$ looks like $\tfrac00$ at $0$ but the curve heads smoothly to $1$ — L'Hospital finds that value fast.
L'Hospital's Rule uses the quotient rule $\dfrac{f'g-fg'}{g^2}$ on the numerator and denominator.
It differentiates $f$ and $g$ separately: $\tfrac{f'}{g'}$, not the quotient rule.
Evaluate $\displaystyle\lim_{x\to0}\dfrac{e^x-1}{x}$ using the rule.
$\tfrac00\Rightarrow\lim\tfrac{e^x}{1}=e^0=1$.
By L'Hospital, $\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=\lim_{x\to0}\dfrac{\cos x}{1}=$
$\cos 0=1$.
Repeat if it's still indeterminate
- If the new ratio $\tfrac{f'}{g'}$ is also $\tfrac00$ or $\tfrac\infty\infty$, apply the rule again.
- Keep differentiating top and bottom until you get a determinate value.
- $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{\sin x}{2x}=\lim_{x\to0}\frac{\cos x}{2}=\tfrac12$ (two applications).
- Each round must still be indeterminate to justify another application.
Applying the rule twice, $\displaystyle\lim_{x\to0}\dfrac{1-\cos x}{x^2}=$ ? (a decimal)
$\to\tfrac{\sin x}{2x}\to\tfrac{\cos x}{2}=\tfrac12$.
$\dfrac{f'}{g'}$ is not the quotient rule. L'Hospital differentiates numerator and denominator separately: $\frac{d}{dx}[f]$ over $\frac{d}{dx}[g]$ — no $\frac{f'g-fg'}{g^2}$. And only use it on $\tfrac00$ or $\tfrac{\infty}{\infty}$; applying it when the limit is already determinate produces nonsense.
Evaluate $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}$.
- Substitute: $\frac{e^0-1}{0}=\frac{0}{0}$ — indeterminate, so L'Hospital applies.
- Differentiate top and bottom: $\dfrac{e^x}{1}$.
- $\displaystyle\lim_{x\to0}\frac{e^x}{1}=e^0=1$.
L'Hospital's Rule: for the indeterminate forms $\tfrac00$ or $\tfrac{\infty}{\infty}$, $\lim\frac{f}{g}=\lim\frac{f'}{g'}$ — differentiate top and bottom separately (not the quotient rule) and re-evaluate. Confirm the form first, and repeat the rule while the result stays indeterminate.