L’Hospital’s Rule and Indeterminate Forms — Question 1

Question 1 - Solution

(a) As $x \to 0$:

- $\sin(3x) \to 0$, and $x \to 0$ - So, the limit is of the indeterminate form $\frac{0}{0}$

(b) Apply L’Hospital’s Rule:

$$\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{d}{dx}[\sin(3x)] \Big/ \frac{d}{dx}[x]$$

$$= \lim_{x \to 0} \frac{3\cos(3x)}{1} = 3\cos(0) = 3(1) = \boxed{3}$$