Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:\mathbb{R}\to (0,\infty )$ be 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

Welcome, fellow traveler of the mathematical landscape. Today, we are going to unravel a problem that might look intimidating at first glance, but beneath its surface lies a beautiful, elegant truth.

We are given a function $f:\mathbb{R}\to (0,\infty )$ that is strictly increasing. This tells us something vital: as we walk along the $x$-axis, the function's value never dips, and it never stays flat. It is always climbing, always reaching for higher ground, and it always stays above the $x$-axis.

Let us visualize the inputs. We are interested in the behavior of $f(x)$ as $x$ approaches infinity.

Consider three points on the $x$-axis: $x$, $5x$, and $7x$. For any positive $x$, it is undeniable that $x<5x<7x$.

Because our function $f$ is strictly increasing, the output values must respect this order. If the input is larger, the output must be larger. Therefore, we can confidently state that $f(x)<f(5x)<f(7x)$. This is our starting point—a simple, geometric truth that will guide us to the solution.

Our goal is to understand the ratio $\frac{f(5x)}{f(x)}$. To get there, let us take our inequality $f(x)<f(5x)<f(7x)$ and divide every term by $f(x)$.

Since we know $f(x)>0$, we do not have to worry about flipping any inequality signs. The expression becomes:

Look at what we have created! We have trapped our target ratio, $\frac{f(5x)}{f(x)}$, between two boundaries: the constant $1$ on the left and the ratio $\frac{f(7x)}{f(x)}$ on the right.

Now, we invite the Squeeze Theorem to the stage. This theorem is the ultimate arbiter in limit problems.

It tells us that if we can show the left side and the right side of our inequality approach the same value as $x\to \infty$, then the middle term has no choice but to follow them. We are given that ${\lim }_{x\to \infty }\frac{f(7x)}{f(x)}=1$.

As we take the limit as $x\to \infty$, our inequality transforms into:

By the Squeeze Theorem, the limit of our middle term must be exactly $1$. The function is squeezed, pinned down by the constraints of the problem.

We have arrived at the final act. The question asks for the value of ${\lim }_{x\to \infty }\left[\frac{f(5x)}{f(x)}-1\right]$.

Since we have established that ${\lim }_{x\to \infty }\frac{f(5x)}{f(x)}=1$, we can simply substitute this value into our expression:

And there it is. The complexity dissolves into a simple, satisfying zero. You have navigated the logic, respected the constraints, and arrived at the truth.