In Exercises 17-36, estimate the limit numerically or state that the limit does not exist. If infinite, state whether the one-sided limits are $\infty$ or $-\infty$
$\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}$

Step 1 ：The limit of a function as x approaches a certain value can be calculated by substituting the value into the function. However, in this case, if we substitute 0 into the function, we get an indeterminate form (0/0). Therefore, we need to use L'Hopital's rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

Step 2 ：The function is \(f = \frac{\sin(2x)}{x}\).

Step 3 ：The derivative of the numerator is \(2\cos(2x)\) and the derivative of the denominator is 1.

Step 4 ：Applying L'Hopital's rule, we get the limit as \(\lim_{x \rightarrow 0} \frac{2\cos(2x)}{1}\).

Step 5 ：Substituting 0 into the function, we get 2.

Step 6 ：Therefore, the limit of the function as x approaches 0 is \(\boxed{2}\).