Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a differentiable function such that \( f^\prime \) is continuous on \( \mathbb{R} \) with \( f^\prime(3) = 18 \). Define \[ g_n(x) = n \left( f \left( x + \frac{5}{n} \right) - f \left( x - \frac{2}{n} \right) \right). \] Then \( \lim_{n \to \infty} g_n(3) \) equals ............... 
 

Hint:

When dealing with expressions involving finite differences, recognize that they often approximate derivatives. Use the derivative's definition to simplify the expression.

Answer 1

Correct Answer: 125 - 127

Step 1: Understand the expression for \( g_n(x) \). 
We are given the function \( g_n(x) = n \left( f \left( x + \frac{5}{n} \right) - f \left( x - \frac{2}{n} \right) \right) \).

Step 2: Use the definition of the derivative. 
The expression for \( g_n(x) \) resembles a finite difference approximation to the derivative of \( f(x) \). We can rewrite the terms inside the parentheses: \[ f \left( x + \frac{5}{n} \right) - f \left( x - \frac{2}{n} \right) \approx f^\prime(x) \left( \frac{5}{n} + \frac{2}{n} \right) = f^\prime(x) \cdot \frac{7}{n}. \]

Step 3: Simplify the expression for \( g_n(x) \). 
Thus, we have: \[ g_n(x) = n \cdot f^\prime(x) \cdot \frac{7}{n} = 7 f^\prime(x). \]

Step 4: Take the limit as \( n \to \infty \). 
Since \( g_n(x) = 7 f^\prime(x) \), and \( f^\prime(3) = 18 \), we find: \[ \lim_{n \to \infty} g_n(3) = 7 \times 18 = 126. \]