Key Concepts and Formulas

• Squeeze Theorem (Sandwich Theorem): If $g(x) \le h(x) \le k(x)$ for all $x$ in an interval containing $c$ (except possibly at $c$), and $\lim_{x \rightarrow c} g(x) = L$ and $\lim_{x \rightarrow c} k(x) = L$, then $\lim_{x \rightarrow c} h(x) = L$.
• Properties of Strictly Increasing Functions: If $f(x)$ is strictly increasing, then $a < b$ implies $f(a) < f(b)$.
• Limit of a Constant: $\lim_{x \rightarrow c} a = a$, where $a$ is a constant.

Step-by-Step Solution

Step 1: Understand the Given Information and Objective

We are given a strictly increasing function $f: \mathbb{R} \rightarrow (0, \infty)$ and that $\lim_{x \rightarrow \infty} \frac{f(7x)}{f(x)} = 1$. Our goal is to find the value of $\lim_{x \rightarrow \infty}\left[\frac{f(5x)}{f(x)}-1\right]$. The codomain $(0, \infty)$ ensures that $f(x) > 0$ for all $x$, which is important for dividing by $f(x)$ later without changing inequality signs.

Step 2: Establish Inequalities Using the Strictly Increasing Property

Since $f(x)$ is strictly increasing, if $a < b$, then $f(a) < f(b)$. As $x$ approaches infinity, consider the relationship between $x$, $5x$, and $7x$. We have $x < 5x < 7x$ for $x > 0$. Applying the strictly increasing function $f$ to these values yields:

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

This inequality holds because $f$ is strictly increasing. This allows us to relate the values of the function at different arguments.

Step 3: Transform the Inequality into Desired Ratios

To obtain the ratios needed for the target limit, divide all parts of the inequality by $f(x)$. Since $f(x) > 0$, the inequality signs are preserved:

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

Simplifying the leftmost term, we get:

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

This step creates the desired ratio $\frac{f(5x)}{f(x)}$ and places it between two expressions for which we know the limits.

Step 4: Apply Limits and the Squeeze Theorem

Now, take the limit as $x$ approaches infinity of all parts of the inequality:

$\lim_{x \rightarrow \infty} 1 \le \lim_{x \rightarrow \infty} \frac{f(5x)}{f(x)} \le \lim_{x \rightarrow \infty} \frac{f(7x)}{f(x)}$

We know that $\lim_{x \rightarrow \infty} 1 = 1$ and $\lim_{x \rightarrow \infty} \frac{f(7x)}{f(x)} = 1$. Substituting these values gives:

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

By the Squeeze Theorem, since $\lim_{x \rightarrow \infty} \frac{f(5x)}{f(x)}$ is bounded between 1 and 1, we have:

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

Step 5: Calculate the Final Limit

Finally, calculate the target limit:

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

Substituting the known limits:

$= 1 - 1 = 0$

Therefore, the limit is 0.

Common Mistakes & Tips

• Assuming a Specific Form for f(x): Avoid assuming $f(x)$ is a polynomial or exponential. The problem is designed to be solved using the given properties (strictly increasing and positive-valued).
• Incorrectly Applying the Squeeze Theorem: Make sure the function is bounded between two functions that converge to the same limit.
• Ignoring the Positive Function Condition: The condition $f: \mathbb{R} \rightarrow (0, \infty)$ is crucial. Without it, dividing by $f(x)$ would be problematic if $f(x)$ could be zero or negative.

Summary

By leveraging the strictly increasing property of the function $f(x)$ and the given limit, we were able to establish inequalities and apply the Squeeze Theorem. This allowed us to determine that $\lim_{x \rightarrow \infty} \frac{f(5x)}{f(x)} = 1$, which then led to the final answer of 0.

The final answer is $\boxed{0}$, which corresponds to option (A).