1. **Problem statement:** Find the limits of the following functions as $x \to 0$: (i) $\lim_{x \to 0} \frac{(1+x)^n - 1}{x}$ (ii) $\lim_{x \to 0} \frac{e^{3x} - 1}{x}$ (iii) $\lim_{x \to 0} \frac{\sin 3x}{5x}$ --- 2. **Formula and rules:** - For small $x$, $(1+x)^n \approx 1 + nx$ (Binomial approximation). - The derivative of $e^{kx}$ at $x=0$ is $k e^{0} = k$. - The limit $\lim_{x \to 0} \frac{\sin ax}{x} = a$. --- 3. **Step-by-step solutions:** (i) Using the binomial approximation: $$ (1+x)^n - 1 \approx 1 + nx - 1 = nx $$ So, $$ \lim_{x \to 0} \frac{(1+x)^n - 1}{x} = \lim_{x \to 0} \frac{nx}{x} = n $$ (ii) Using the exponential limit: $$ \lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{e^{3x} - e^0}{x} $$ This is the definition of the derivative of $e^{3x}$ at $x=0$: $$ = \left. \frac{d}{dx} e^{3x} \right|_{x=0} = 3 e^{0} = 3 $$ (iii) Using the sine limit property: $$ \lim_{x \to 0} \frac{\sin 3x}{5x} = \frac{1}{5} \lim_{x \to 0} \frac{\sin 3x}{x} = \frac{1}{5} \times 3 = \frac{3}{5} $$ --- **Final answers:** (i) $n$ (ii) $3$ (iii) $\frac{3}{5}$