Question

If \( f:\mathbb{R} \to (0,\infty) \) is an increasing function and if \( \lim_{x \to 2018} \frac{f(3x)}{f(x)} = 1 \), then \( \lim_{x \to 2018} \frac{f(2x)}{f(x)} \) is equal to

• A. \( \frac{2}{3} \)
• B. \( \frac{3}{2} \)
• C. \( 2 \)
• D. \( 3 \)
• E. \( 1 \)

Hint:

For increasing functions, ratios approaching 1 indicate very slow growth near that point.

Answer 1

Answer: (E)

Concept: Given \( f \) is increasing and positive. If: \[ \lim_{x \to a} \frac{f(kx)}{f(x)} = 1 \] it suggests that function behaves almost like a constant near that point.

Step 1: Given: \[ \lim_{x \to 2018} \frac{f(3x)}{f(x)} = 1 \]

Step 2: Since function is increasing, we use squeeze-type reasoning.
As \( x \to 2018 \), both \( 2x \) and \( 3x \) approach finite values.

Step 3: Using behavior: \[ \frac{f(2x)}{f(x)} \to 1 \]

Step 4: Conclude: \[ = 1 \]