We are tasked with finding $\lim_{{x \to \infty}} f(x)$, given that for all $x > 1$, the function $f(x)$ satisfies the inequality:

$\frac{10e^x - 23}{2e^x} < f(x) < \frac{5\sqrt{x}}{\sqrt{x-1}}.$

Step 1: Investigate the left-hand side limit as $x \to \infty$

The left-hand side expression is:

$\frac{10e^x - 23}{2e^x} = \frac{10e^x}{2e^x} - \frac{23}{2e^x} = 5 - \frac{23}{2e^x}.$

As $x \to \infty$, the term $\frac{23}{2e^x} \to 0$ because $e^x$ grows exponentially. Therefore:

$\lim_{{x \to \infty}} \left( 5 - \frac{23}{2e^x} \right) = 5.$

Step 2: Investigate the right-hand side limit as $x \to \infty$

Now let's examine the right-hand side expression:

$\frac{5\sqrt{x}}{\sqrt{x-1}}.$

For large $x$, we can approximate $\sqrt{x - 1} \approx \sqrt{x}$. Hence:

$\frac{5\sqrt{x}}{\sqrt{x-1}} \approx \frac{5\sqrt{x}}{\sqrt{x}} = 5.$

Thus, as $x \to \infty$:

$\lim_{{x \to \infty}} \frac{5\sqrt{x}}{\sqrt{x-1}} = 5.$

Step 3: Apply the Squeeze Theorem

We now have the following situation as $x \to \infty$:

$5 \leq f(x) \leq 5.$

By the Squeeze Theorem, since both bounds approach 5, we conclude that:

$\lim_{{x \to \infty}} f(x) = 5.$

Thus, the limit of $f(x)$ as $x \to \infty$ is 5.

Would you like more details on any step, or do you have any further questions?

Here are five related questions to explore:

1. What is the Squeeze Theorem, and how does it apply in this case?
2. How do exponential functions behave as $x \to \infty$?
3. How do you approximate square roots of large numbers, like $\sqrt{x - 1}$?
4. Can you apply the Squeeze Theorem to other types of inequalities?
5. What would happen if the limits of the upper and lower bounds were not equal?

Tip: Always check the behavior of functions like exponentials and roots as $x \to \infty$, as they often simplify in the limit.