If x ( f(7x)f(x)) = 1, then x [ ( f(5x)f(x)) - 1 ] is: — Solution | JEE Mathematics

MCQEasyJEE 2024Limits

If $\lim_{x \to \infty} \left(\frac{f(7x)}{f(x)}\right) = 1$ , then $\lim_{x \to \infty} \left[\left(\frac{f(5x)}{f(x)}\right) - 1\right]$ is:

Answer

Correct answer:B

Standard Method

Given: $\lim_{x \to \infty} \frac{f(7x)}{f(x)} = 1$ and $f$ is strictly increasing.

Find: $\lim_{x \to \infty} \left[\frac{f(5x)}{f(x)} - 1\right]$

Since $f$ is strictly increasing and $x < 5x < 7x$ for large $x$ , we have

f(x) < f(5x) < f(7x)

Dividing throughout by $f(x)$ ,

1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}

Now use the given limit

\lim_{x \to \infty} \frac{f(7x)}{f(x)} = 1

So by the squeeze theorem,

\lim_{x \to \infty} \frac{f(5x)}{f(x)} = 1

Therefore,

\lim_{x \to \infty} \left[\frac{f(5x)}{f(x)} - 1\right] = 1 - 1 = 0

Hence, the value of the limit is $0$ . The correct option is B.

Using monotonicity and squeezing

Given: $\lim_{x \to \infty} \frac{f(7x)}{f(x)} = 1$ .

Find: $\lim_{x \to \infty} \left[\frac{f(5x)}{f(x)} - 1\right]$

For sufficiently large positive $x$ ,

x < 5x < 7x

Because $f$ is strictly increasing,

f(x) < f(5x) < f(7x)

Divide each term by $f(x)$ to get

1 < \frac{f(5x)}{f(x)} < \frac{f(7x)}{f(x)}

Taking limit as $x \to \infty$ ,

1 \le \lim_{x \to \infty} \frac{f(5x)}{f(x)} \le 1

Thus,

\lim_{x \to \infty} \frac{f(5x)}{f(x)} = 1

Now subtract $1$ :

\lim_{x \to \infty} \left[\frac{f(5x)}{f(x)} - 1\right] = 0

Therefore, the correct option is B.

Assuming directly that $\lim_{x \to \infty} \frac{f(cx)}{f(x)} = 1$ for every constant $c$ from the single condition at $c = 7$ is not justified in general. Here, the correct step is to use the fact that $f$ is strictly increasing and apply the squeeze theorem with $x < 5x < 7x$ .
Forgetting to use the monotonicity of $f$ is a conceptual error. The information that $f$ is strictly increasing gives the inequality $f(x) < f(5x) < f(7x)$ , which is the key bridge from the given limit to the required one.
Subtracting $1$ before establishing $\lim_{x \to \infty} \frac{f(5x)}{f(x)}$ can lead to circular reasoning. First find the limit of the ratio using squeezing, then evaluate $\lim_{x \to \infty} \left[\frac{f(5x)}{f(x)} - 1\right]$ .

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