L’Hospital’s Rule and Indeterminate Forms — Question 10

Question 10 - Solution

As $x \to 0$: $$\sin(5x) \to 0, \quad \tan(3x) \to 0 \Rightarrow \text{Indeterminate form } \frac{0}{0}$$

Apply L’Hospital’s Rule: $$\lim_{x \to 0} \frac{\sin(5x)}{\tan(3x)} = \lim_{x \to 0} \frac{d}{dx} \left[ \frac{\sin(5x)}{\tan(3x)} \right] = \lim_{x \to 0} \frac{5 \cos(5x)}{3 \sec^2(3x)}$$

$$= \lim_{x \to 0} \frac{5 \cos(5x)}{3 \cdot \frac{1}{\cos^2(3x)}} = \lim_{x \to 0} \frac{5 \cos(5x) \cos^2(3x)}{3}$$

As $x \to 0$: $$\cos(5x) \to 1, \quad \cos(3x) \to 1 \Rightarrow \cos^2(3x) \to 1$$

So, $$\frac{5 \cdot 1 \cdot 1}{3} = \frac{5}{3}$$

Final Answer: $\boxed{\dfrac{5}{3}}$