To evaluate the limit
[tex]\[
\lim_{x \rightarrow 0} \frac{3 \sin 8x}{5 x}
\][/tex]
we will use l'Hôpital's Rule because the expression [tex]\(\frac{0}{0}\)[/tex] is of indeterminate form as [tex]\(x \rightarrow 0\)[/tex]. L'Hôpital's Rule tells us that if the limits of the numerator and denominator both tend to zero, the limit of the quotient can be found by differentiating the numerator and the denominator.
First, let's apply l'Hôpital's Rule which requires us to differentiate the numerator and the denominator.
The numerator is [tex]\(3 \sin 8x\)[/tex]:
[tex]\[
\frac{d}{dx}[3 \sin 8x] = 3 \cdot (8 \cos 8x) = 24 \cos 8x
\][/tex]
The denominator is [tex]\(5x\)[/tex]:
[tex]\[
\frac{d}{dx}[5x] = 5
\][/tex]
Using l'Hôpital's Rule, we can rewrite the original limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{3 \sin 8x}{5 x} = \lim_{x \rightarrow 0} \frac{24 \cos 8x}{5}
\][/tex]
Now, we evaluate the limit of this new expression:
[tex]\[
\lim_{x \rightarrow 0} \frac{24 \cos 8x}{5}
\][/tex]
We know that [tex]\(\cos 8x \rightarrow \cos 0 = 1\)[/tex] as [tex]\(x \rightarrow 0\)[/tex]. Therefore:
[tex]\[
\frac{24 \cos 8 x}{5} \rightarrow \frac{24 \cdot 1}{5} = \frac{24}{5}
\][/tex]
Hence, the limit is:
[tex]\[
\lim_{x \rightarrow 0} \frac{3 \sin 8x}{5 x} = \frac{24}{5}
\][/tex]
So, the exact answer is:
[tex]\[
\frac{24}{5}
\][/tex]
