To solve the limit $\lim_{x \to 0} \frac{\sin(2x)}{x}$, let's proceed step-by-step using limit properties and trigonometric identities.

Step-by-Step Solution

1. Use Substitution: Let's make a substitution to help simplify the limit. Set $u = 2x$. As $x \to 0$, $u \to 0$ as well.

   Rewrite the expression in terms of $u$: $\lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{u \to 0} \frac{\sin(u)}{\frac{u}{2}}$

2. Simplify the Expression: The expression $\frac{\sin(u)}{\frac{u}{2}}$ can be simplified by multiplying the numerator and the denominator by 2: $\lim_{u \to 0} \frac{\sin(u)}{\frac{u}{2}} = \lim_{u \to 0} \frac{2\sin(u)}{u}$

3. Apply the Sine Limit Identity: We know from the standard limit identity that $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$.

   Applying this identity: $\lim_{u \to 0} \frac{2\sin(u)}{u} = 2 \lim_{u \to 0} \frac{\sin(u)}{u} = 2 \times 1 = 2$

Conclusion

Thus, the limit is: $\lim_{x \to 0} \frac{\sin(2x)}{x} = 2$

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

Here are some related questions to further your understanding:

1. What is the limit of $\lim_{x \to 0} \frac{\sin(3x)}{x}$?
2. How does the substitution method help in evaluating limits involving trigonometric functions?
3. Can we apply L'Hopital's Rule to solve $\lim_{x \to 0} \frac{\sin(2x)}{x}$? What would be the result?
4. What are some other common limits involving trigonometric functions?
5. How would the solution change if the function was $\lim_{x \to 0} \frac{\tan(2x)}{x}$?

Tip: Always consider using trigonometric identities and substitution when simplifying limits involving trigonometric functions.