Evaluate the limit. $$ \lim _{x \rightarrow \infty} x^{3}\left(\frac{2}{x}-\sin \frac{2}{x}\right)=\square $$ (Type an exact answer.)

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To evaluate the limit:

$$
\lim_{x \rightarrow \infty} x^{3}\left(\frac{2}{x} - \sin \frac{2}{x}\right)
$$

Let's make a substitution to simplify the expression. Let $$ t = \frac{1}{x} $$. As $$ x \rightarrow \infty $$, $$ t \rightarrow 0^+ $$. Substituting, the limit becomes:

$$
\lim_{t \rightarrow 0^+} \frac{2t - \sin(2t)}{t^3}
$$

Next, we use the Taylor series expansion for $$ \sin(2t) $$ around $$ t = 0 $$:

$$
\sin(2t) = 2t - \frac{(2t)^3}{6} + \frac{(2t)^5}{120} - \cdots
$$

Subtracting $$ \sin(2t) $$ from $$ 2t $$:

$$
2t - \sin(2t) = 2t - \left(2t - \frac{8t^3}{6} + \cdots\right) = \frac{4}{3} t^3 + \cdots
$$

Now, divide by $$ t^3 $$:

$$
\frac{2t - \sin(2t)}{t^3} = \frac{\frac{4}{3} t^3 + \cdots}{t^3} = \frac{4}{3} + \cdots
$$

Taking the limit as $$ t \rightarrow 0 $$:

$$
\lim_{t \rightarrow 0^+} \frac{2t - \sin(2t)}{t^3} = \frac{4}{3}
$$

Thus, the original limit is:

$$
\frac{4}{3}
$$

**Answer:** $$\dfrac{4}{3}$$
The limit is $$\frac{4}{3}$$.

Evaluate the limit.
(Type an exact answer.)

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