https://file.aiquickdraw.com/mathbot/file/1730340654450fj16y6vv.jpg

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.

Evaluate the limit of (sin(5x) / x) as x approaches 0.

Explore the solution to finding the limit of (sin(5x) / x) as x approaches 0. This solution applies a fundamental limit property for trigonometric functions, demonstrating how the limit results in the constant value of 5. Ideal for students in Grades 11-12 learning advanced calculus concepts.

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