Mathematics - Limits Question with Solution | TestHub

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

Options:

A. $4$

B. $0$

C. $\frac{7}{5}$

D. $1$

Correct Answer: B

Solution:

Given that $f:R\to (0,\infty )$ is a strictly increasing function. This means that for any $a<b$, we have $f(a)<f(b)$.

Also, we are given the limit:

$$\underset{x\to \infty }{\lim }\frac{f(7x)}{f(x)}=1$$

We need to find the value of ${\lim }_{x\to \infty }\left[\frac{f(5x)}{f(x)}-1\right]$.

Consider a positive real number $x$. Since $x\to \infty$, we can assume $x>0$.

For $x>0$, we have the inequality:

$$x<5x<7x$$

Since $f$ is a strictly increasing function, applying $f$ to this inequality preserves the order:

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

Since $f(x)\in (0,\infty )$, $f(x)$ is always positive. We can divide the entire inequality by $f(x)$ without changing the direction of the inequalities:

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

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

Now, we take the limit as $x\to \infty$ for all parts of the inequality:

$$\underset{x\to \infty }{\lim }1<\underset{x\to \infty }{\lim }\frac{f(5x)}{f(x)}<\underset{x\to \infty }{\lim }\frac{f(7x)}{f(x)}$$

We know that ${\lim }_{x\to \infty }1=1$.

We are given that ${\lim }_{x\to \infty }\frac{f(7x)}{f(x)}=1$.

So, the inequality becomes:

$$1<\underset{x\to \infty }{\lim }\frac{f(5x)}{f(x)}<1$$

By the Squeeze Theorem (also known as the Sandwich Theorem), if a function is bounded between two other functions that converge to the same limit, then the function itself must converge to that limit.

In this case, $\frac{f(5x)}{f(x)}$ is "squeezed" between $1$ and $\frac{f(7x)}{f(x)}$, both of which approach $1$ as $x\to \infty$.

Therefore, we can conclude that:

$$\underset{x\to \infty }{\lim }\frac{f(5x)}{f(x)}=1$$

Finally, we need to find the value of ${\lim }_{x\to \infty }\left[\frac{f(5x)}{f(x)}-1\right]$.

Using the properties of limits, we can write this as:

$$\underset{x\to \infty }{\lim }\left[\frac{f(5x)}{f(x)}-1\right]=\underset{x\to \infty }{\lim }\frac{f(5x)}{f(x)}-\underset{x\to \infty }{\lim }1$$

$$=1-1$$

$$=0$$

The final answer is $\boxed{0}$.