Let's solve the given limit step by step: \[ \lim_{\Delta x \to 0^+} \frac{\frac{3}{x+\Delta x} - \frac{3}{x}}{\Delta x} \] This is the definition of the derivative of the function \( f(x) = \frac{3}{x} \). --- ### Step 1: Combine the Numerator Combine the two fractions in the numerator: \[ \frac{3}{x+\Delta x} - \frac{3}{x} = 3 \left( \frac{1}{x+\Delta x} - \frac{1}{x} \right) \] Find a common denominator: \[ \frac{1}{x+\Delta x} - \frac{1}{x} = \frac{x - (x+\Delta x)}{x(x+\Delta x)} = \frac{-\Delta x}{x(x+\Delta x)} \] So the numerator becomes: \[ 3 \left( \frac{-\Delta x}{x(x+\Delta x)} \right) = \frac{-3\Delta x}{x(x+\Delta x)} \] --- ### Step 2: Substitute Back into the Limit \[ \lim_{\Delta x \to 0^+} \frac{\frac{-3\Delta x}{x(x+\Delta x)}}{\Delta x} \] This simplifies to: \[ \lim_{\Delta x \to 0^+} \frac{-3\Delta x}{x(x+\Delta x) \Delta x} \] --- ### Step 3: Simplify the Expression Cancel \(\Delta x\) in the numerator and denominator: \[ \frac{-3\Delta x}{x(x+\Delta x)\Delta x} = \frac{-3}{x(x+\Delta x)} \] --- ### Step 4: Take the Limit Let \(\Delta x \to 0^+\): \[ \lim_{\Delta x \to 0^+} \frac{-3}{x(x+\Delta x)} = \frac{-3}{x^2} \] --- ## **Final Answer** \[ \boxed{\frac{-3}{x^2}} \] This is the derivative of \( f(x) = \frac{3}{x} \) with respect to \( x \).

The limit represents the derivative of \(f(x) = \frac{3}{x}\), calculated using the definition of the derivative as \(\lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}\). Simplifying the difference quotient yields \(\frac{-3}{x^2}\), the derivative of the function.