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

Question

Let f:→R→(0,∞) be strictly increasing function such that limxf(7x)f(x)=1 \lim_{x \to \infty} f(7x)f(x) = 1 . Then, the value of limx[f(5x)f(x)1] \lim_{x \to \infty}[f(5x)f(x)−1] is equal to

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Solution

The problem is asking for the limit of a function as x approaches infinity. Here are the steps to solve it:

  1. Given that lim(x→∞) [f(7x) / f(x)] = 1, we can infer that f(7x) ~ f(x) as x→∞. This is because the limit of the ratio of two functions is 1, which implies that the two functions are approximately equal for large x.

  2. We need to find the value of lim(x→∞) [f(5x) / f(x) - 1]. We can rewrite this as lim(x→∞) [f(5x) / f(x)] - lim(x→∞) 1.

  3. Since f(x) is a strictly increasing function, we can infer that f(5x) > f(x) for all x > 0. Therefore, the limit of the ratio f(5x) / f(x) as x approaches infinity should be greater than 1.

  4. However, we don't have enough information to determine the exact value of this limit. We know that it's greater than 1, but without more information about the function f(x), we can't find a specific numerical value.

  5. Therefore, the value of lim(x→∞) [f(5x) / f(x) - 1] is indeterminate based on the information given in the problem.

This problem has been solved

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