L’HAZpital’s Rule for Indeterminate Limits

3.2 L’HÃŽpital’s Rule for Indeterminate Limits

In this section, we learn a powerful method to attack problems on indeterminate forms, called l’HÃŽpital’s rule (also written l’Hospital’s rule).

L’HÃŽpital’s Rule for the Indeterminate Forms \(0/0\) and \(\infty /\infty \)

Assume \(f\) and \(g\) are two functions with \(f(a)=g(a)=0\). Then for \(x\neq a\), we have \[ \frac {f(x)}{g(x)}=\frac {f(x)-\overbrace {f(a)}^{0}}{g(x)-\underbrace {g(a)}_{0}}=\frac {\dfrac {f(x)-f(a)}{x-a}}{\dfrac {g(x)-g(a)}{x-a}} \] Suppose the derivatives \(f'(a)\) and \(g'(a)\) exist and \(g'(a)\neq 0\). Because \[ \lim _{x\to a}\frac {f(x)-f(a)}{x-a}=\lim _{h\to 0}\frac {f(a+h)-f(a)}{h}=f'(a),\qquad (h=x-a) \] and \[ \lim _{x\to a}\frac {g(x)-g(a)}{x-a}=g'(a), \] we get \[ \boxed {\lim _{x\to a}\frac {f(x)}{g(x)}=\frac {f'(a)}{g'(a)}.\tag {i}} \]

The following theorem generalizes the cases where we can use Equation (i). It states that Equation (i) is true under less stringent conditions and is also valid for one-sided limits as well as limits at \(+\infty \) and \(-\infty \).

Exercise 5. Find \(\displaystyle \lim _{x\to 0}\frac {\sin x-\tan x}{x^{3}}.\)