JEE Main 2024 · 31 January · evening · Single correct · +4 / −1

Let f: → ℝ →(0, ∞) be strictly increasing function such that lim x → ∞ f(7 x)/f(x)=1. Then, the value of lim x → ∞[f(5 x)/f(x)-1] is equal to

  1. (a) 0
  2. (b) 4
  3. (c) 1
  4. (d) 7/5
Answer and solution

Answer (a) 0

Solution

Maths sir · 4 steps

Since f is increasing, f(5x) sits between f(x) and f(7x), so the 5x ratio is trapped between 1 and 1.

  1. Order the values. For large positive x we have x < 5x < 7x. Since f is strictly increasing, f(x) < f(5x) < f(7x).
  2. Divide by f(x). f(x) > 0, so dividing by it keeps the inequalities the same way round.

    1 < f(5x)/f(x) < f(7x)/f(x)

  3. Sandwich. As x → ∞ the left side is always 1 and the right side tends to 1. The middle ratio is squeezed, so lim(x → ∞) f(5x)/f(x) = 1.
  4. So. The required limit is 1 - 1 = 0, which is option (a).

The idea to keep. When you know a limit only at one point like 7x, use monotonicity to trap the unknown expression between known ones, then apply the sandwich theorem.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2024.