Let $f: \mathbb{R} \rightarrow (0, \infty)$ be strictly increasing function such that $\lim_{x \rightarrow \infty} \frac{f(7x)}{f(x)}=1$. Then, the value of $\lim_{x \rightarrow \infty}\left[\frac{f(5x)}{f(x)}-1\right]$ is equal to

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Limits

Let

f : R \to (0, \infty)

be strictly increasing function such that

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

. Then, the value of

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

is equal to

Answer: (B)

For

x > 0

, since

x < 5 x < 7 x

and

f

is strictly increasing on

R

f (x) \leq f (5 x) \leq f (7 x)

.

Since

f

is positive, divide through by

f (x) > 0

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

.

As

x \to \infty

\frac{f ( 7 x )}{f ( x )} \to 1

, so by the squeeze theorem,

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

.

Hence

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

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option

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