To find the limit [tex]\(\lim_{x \rightarrow 0} \frac{\sin 5x}{\sin 2x}\)[/tex], we can follow these steps:
1. Expression of Sine Functions:
The expression we are dealing with is [tex]\(\frac{\sin 5x}{\sin 2x}\)[/tex].
2. Behavior Near Zero:
As [tex]\(x\)[/tex] approaches 0, both [tex]\(\sin 5x\)[/tex] and [tex]\(\sin 2x\)[/tex] approach 0. To find the limit, we can use the fact that [tex]\(\sin(kx) \approx kx\)[/tex] when [tex]\(x\)[/tex] is near 0, where [tex]\(k\)[/tex] is a constant. Thus, [tex]\(\sin 5x \approx 5x\)[/tex] and [tex]\(\sin 2x \approx 2x\)[/tex].
3. Substituting Approximation:
Now, substitute these approximations into the original limit expression:
[tex]\[
\lim_{x \rightarrow 0} \frac{\sin 5x}{\sin 2x} \approx \lim_{x \rightarrow 0} \frac{5x}{2x}
\][/tex]
4. Simplifying the Fraction:
Simplify the expression by cancelling out [tex]\(x\)[/tex] in the numerator and the denominator:
[tex]\[
\frac{5x}{2x} = \frac{5}{2}
\][/tex]
5. Conclusion:
Therefore, the limit is:
[tex]\[
\lim_{x \rightarrow 0} \frac{\sin 5x}{\sin 2x} = \frac{5}{2}
\][/tex]
So, the final answer is [tex]\(\frac{5}{2}\)[/tex].
