The expression in the image is:

$\frac{\sin(5x)}{x}$

If you're interested in evaluating the limit of this expression as $x \to 0$, a common technique would be to use the standard limit property: $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$ for any constant $k$. Here, $k = 5$, so:

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

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

Here are some related questions to explore further:

1. What happens if we take the limit of $\frac{\sin(kx)}{x}$ as $x \to \infty$?
2. How would this limit change if the argument of the sine function was $5x^2$ instead of $5x$?
3. Can this limit be evaluated using L'Hôpital's Rule instead?
4. How does the small-angle approximation for sine help us understand limits like this?
5. What other trigonometric limits are similar to $\frac{\sin(kx)}{x}$ as $x \to 0$?

Tip: For trigonometric limits involving sine or cosine, rewriting the function in terms of standard forms can make the limit evaluation much simpler.