To find the limit of \(\frac{\sin(5x)}{x}\) as \(x \to 0\), we can use the standard limit:

\[
\lim_{x \to 0} \frac{\sin(x)}{x} = 1.
\]

We will rewrite the expression to match the standard form. Let's consider:

\[
\lim_{x \to 0} \frac{\sin(5x)}{x}.
\]

We can manipulate this by multiplying and dividing by 5:

\[
\lim_{x \to 0} \frac{\sin(5x)}{x} = \lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \cdot 5 \right).
\]

Now, we can separate the constant 5 from the limit:

\[
= 5 \cdot \lim_{x \to 0} \frac{\sin(5x)}{5x}.
\]

Next, we recognize that if \(u = 5x\), as \(x \to 0\), \(u \to 0\) as well. Therefore:

\[
\lim_{x \to 0} \frac{\sin(5x)}{5x} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1.
\]

Substitute this result back:

\[
= 5 \cdot 1 = 5.
\]

Thus, the limit is:

\[
\boxed{5}.
\]

Would you like more details on this solution, or have any questions?

Here are five related questions to explore further:

1. What is the limit of \(\frac{\sin(3x)}{x}\) as \(x \to 0\)?
2. How do you find the limit of \(\frac{\tan(x)}{x}\) as \(x \to 0\)?
3. Why is the limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) fundamental in calculus?
4. How would the limit change if the function were \(\frac{\sin(x)}{x^2}\) as \(x \to 0\)?
5. What is the Taylor series expansion of \(\sin(x)\), and how does it relate to limits?

**Tip:** Remember that many trigonometric limits can be solved using small-angle approximations or Taylor series expansions!

Discover how to solve the limit of sin(5x)/x as x approaches 0 using key calculus concepts and the standard trigonometric limit formula. This detailed solution explains each step and is ideal for students in Grades 11-12 or introductory calculus courses.

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