Lâ€™HAZpitalâ€™s Rule for Indeterminate Limits

### 3.2 Lâ€™HÃƒÅ½pitalâ€™s Rule for Indeterminate Limits

In this section, we learn a powerful method to attack problems on indeterminate forms, called lâ€™HÃƒÅ½pitalâ€™s rule (also written lâ€™Hospitalâ€™s rule).

#### Lâ€™HÃƒÅ½pitalâ€™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 $$\displaystyle \lim _{x\to 0}\frac {\sin x-\tan x}{x^{3}}.$$