Sử dụng Quy tắc L'Hospital để xác định giới hạn của các dạng vô định

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.

When f(a) = g(a) = 0 the ratio f/g near a matches the ratio of their slopes

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.

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.