The expression you've provided is:

$\lim_{x \to 10} \frac{2x - 5 \sin(2x)}{x - 5 \sin(x)}$

Since you requested not to use L'Hôpital's rule, we can approach this problem by analyzing the limits of the individual terms and simplifying the expression algebraically if possible.

Let’s first substitute $x = 10$ directly to check if the limit can be evaluated easily.

Substituting $x = 10$ into the expression:

$\frac{2(10) - 5 \sin(2(10))}{10 - 5 \sin(10)}$

We’ll calculate this step-by-step.

1. $2(10) = 20$
2. $5 \sin(20)$
3. $5 \sin(10)$

Now, let me compute those sine values and evaluate the limit.After substituting $x = 10$, the limit evaluates to approximately:

$\frac{2(10) - 5 \sin(20)}{10 - 5 \sin(10)} \approx 1.213$

Thus, the limit is approximately $1.213$, and we were able to compute it without using L'Hôpital's rule by direct substitution.

Would you like further details on this approach, or do you have any other questions?

Here are five related questions you might explore:

1. What happens if we try to evaluate the limit as $x \to 0$ instead?
2. How can we approximate this limit using series expansions?
3. What if we applied L'Hôpital's rule—would we get the same result?
4. How does the behavior of $\sin(x)$ affect this limit when $x$ is large or small?
5. How would this limit change if we used cosine instead of sine?

Tip: If direct substitution doesn't work, consider simplifying the expression or using approximation techniques like Taylor series expansion around critical points.