To find the value of the limit [tex]\(\lim_{x \to 0} \frac{\sin(2x)}{x}\)[/tex], we will go through the problem step by step.
Let's start by examining the expression [tex]\(\frac{\sin(2x)}{x}\)[/tex]. As [tex]\(x\)[/tex] approaches 0, both the numerator [tex]\(\sin(2x)\)[/tex] and the denominator [tex]\(x\)[/tex] approach 0, which is an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. Therefore, we can use L'Hôpital's Rule, which is applicable in such cases, to find the limit. L'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] results in [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\][/tex]
In our case, [tex]\(f(x) = \sin(2x)\)[/tex] and [tex]\(g(x) = x\)[/tex].
First, we find the derivatives of the numerator and the denominator:
- The derivative of [tex]\(\sin(2x)\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx} \sin(2x) = 2 \cos(2x) \][/tex]
- The derivative of [tex]\(x\)[/tex] with respect to [tex]\(x\)[/tex] is:
[tex]\[ g'(x) = \frac{d}{dx} x = 1 \][/tex]
Now, applying L'Hôpital's Rule, we get:
[tex]\[
\lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{2 \cos(2x)}{1}
\][/tex]
Next, let's evaluate the limit of the right-hand side expression as [tex]\(x\)[/tex] approaches 0:
[tex]\[
\lim_{x \to 0} 2 \cos(2x)
\][/tex]
Since [tex]\(\cos(2x)\)[/tex] approaches [tex]\(\cos(0) = 1\)[/tex] as [tex]\(x\)[/tex] approaches 0, this simplifies to:
[tex]\[
2 \times 1 = 2
\][/tex]
Therefore, the value of the limit is:
[tex]\[
\lim_{x \to 0} \frac{\sin(2x)}{x} = 2
\][/tex]
Among the given options:
- 2
- [tex]\(\frac{1}{2}\)[/tex]
- [tex]\(\infty\)[/tex]
The correct answer is:
[tex]\[
2
\][/tex]
