Question

In Exercises \(19 - 28 ,\) determine the limit graphically. Confirm algebraically. $$\lim _ { x \rightarrow 0 } \frac { \sin 2 x } { x }$$

Short answer

The limit of \( \frac { \sin 2 x } { x } \) as \(x\) approaches 0 is 2.

Step-by-step solution

Plot the function

Creating a graph of the function \( \frac { \sin 2 x } { x }\) as \(x\) approaches 0, will give an idea of what the function behaves. If the function seems to be approaching a particular y-value as \(x\) gets close to 0, then that y-value will likely be the limit.

Apply L'Hopital's rule

Since the given limit is of the indeterminate form 0/0 when \(x\) approaches 0, we can use L'Hopital's rule. We need to differentiate the numerator and the denominator separately. The derivative of \(\sin 2 x\) using the chain rule is \(2 \cos 2 x \) and the derivative of \(x\) is 1.

Compute the new limit

Having differentiated the numerator and denominator, our new function becomes \( \frac {2 \cos 2 x}{1} \), which simplifies to \( 2 \cos 2 x \). Then, the limit as \(x\) approaches 0 is \( 2 \cos 0 \) which becomes 2.

Confirm the limit

Comparing the graphical limit and the limit we found algebraically, they confirm each other, so we can now say confidently that as \(x\) approaches 0, the limit of \( \frac { \sin 2 x } { x }\) is 2.