The expression in the image is:

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

To solve this limit, we can use the standard limit:

$\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$

Rewriting the expression in a more convenient form, we can introduce a factor of 3 in the numerator and denominator:

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

Now, using the standard limit mentioned earlier:

$\lim_{{x \to 0}} \frac{\sin(3x)}{3x} = 1$

Substitute this result back:

$\frac{3}{5} \cdot 1 = \frac{3}{5}$

Thus, the limit is:

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

Would you like more details or have any questions?

Further Related Questions:

1. How can we use L'Hopital's Rule to solve this limit?
2. What is the limit of $\frac{\sin(kx)}{mx}$ as $x \to 0$ for any constants $k$ and $m$?
3. How do we evaluate limits involving trigonometric functions?
4. What is the proof of the standard limit $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$?
5. How do we compute limits of the form $\lim_{{x \to a}} \frac{f(x)}{g(x)}$ where both $f(a) = 0$ and $g(a) = 0$?

Tip:

When solving limits involving trigonometric functions, consider rewriting them in forms that make use of standard limits like $\lim_{{x \to 0}} \frac{\sin(x)}{x} = 1$.