The problem asks us to evaluate the following limit:

$\lim_{{x \to 0}} \frac{{\sin(5x)}}{{x(x + 3)}}$

Step 1: Apply limit properties

We first examine the form of the limit as $x \to 0$. Direct substitution of $x = 0$ would give:

$\frac{{\sin(0)}}{{0(0 + 3)}} = \frac{0}{0}$

This is an indeterminate form, so we need to proceed with further analysis.

Step 2: Apply standard limit for small angles

Recall that for small values of $x$, $\sin(x)$ can be approximated by $x$, meaning:

$\lim_{{x \to 0}} \frac{\sin(5x)}{x} = 5$

This is a standard limit result. Now, we need to deal with the remaining part of the expression, which is $\frac{1}{x + 3}$.

Step 3: Rewrite the limit expression

Thus, the original limit can be rewritten as:

$\lim_{{x \to 0}} \frac{\sin(5x)}{x(x + 3)} = \lim_{{x \to 0}} \frac{5}{x + 3}$

Step 4: Substitution of $x = 0$

Now, we substitute $x = 0$ into the remaining expression:

$\frac{5}{0 + 3} = \frac{5}{3}$

Thus, the value of the limit is $\frac{5}{3}$, which corresponds to answer a.

Final Answer:

The correct choice is a. $\frac{5}{3} = 1,(6)$.

Would you like further details or have any questions? Here are 5 related questions you might want to explore:

1. How is the limit of $\frac{\sin(x)}{x}$ derived as $x \to 0$?
2. What are the common indeterminate forms encountered in limits?
3. How do you apply L'Hopital's Rule for limits with indeterminate forms?
4. How do small angle approximations work for trigonometric functions?
5. What other techniques can be used to resolve limits without direct substitution?

Tip: Always check for standard limit results like $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$ when dealing with trigonometric limits.