$$\lim_{x \to 0^+} \frac{\tan(5x^{1/3}) \cdot \log(1+3x)}{5x^{1/3}(\tan^{-1}(3\sqrt{x}))(e^{5x^{4/3}}-1)}$$
Is equal to:
Step-by-Step Solution
Key Concept: Expand the expression near the limiting point using the appropriate standard trigonometric, logarithmic, or exponential approximation.
$$\lim_{x \to 0^+} \frac{\tan(5x^{1/3})}{5x^{1/3}} \cdot \frac{\log(1+3x)}{3x} \cdot \frac{3x}{(\tan^{-1}(3\sqrt{x}))^2} \cdot \frac{(3\sqrt{x})^2}{(\tan^{-1}(3\sqrt{x}))^2} \cdot \frac{5x^{4/3}}{e^{5x^{4/3}}-1}$$
Using standard limits:
- $\lim_{u \to 0} \frac{\tan u}{u} = 1$
- $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$
- $\lim_{u \to 0} \frac{\tan^{-1}u}{u} = 1$
- $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$
$$= 1 \cdot 1 \cdot \frac{3}{1} \cdot \frac{9x}{9x} \cdot 1 = \frac{1}{3}$$
Correct Answer: 3