### Video Transcript

Find the limit as π₯ approaches zero of four sin six π₯ over two π₯.

In order to find a limit, our first instinct might be to use direct substitution. However, if we try to use direct substitution to find the limit given to us in the question, we obtain zero over zero, which is undefined and is an indeterminate form. So we cannot use direct substitution to find the limit given to us in the question.

However, we can use LβHΓ΄pitalβs rule. LβHΓ΄pitalβs rule says that if the limit as π₯ approaches π of π of π₯ over π of π₯ is of the indeterminate form zero over zero, β over β, or negative β over β. Then it is equal to the limit as π₯ approaches π of the first derivative of π with respect to π₯ over the first derivative of π with respect to π₯. Letβs use LβHΓ΄pitalβs rule to find the limit given to us in the question.

Let π of π₯ equal four sin six π₯ and π of π₯ equal two π₯. Letβs find the derivatives of π and π with respect to π₯. We have that, for all constants π and π, the derivative of π sin ππ₯ with respect to π₯ is π times π times cos ππ₯. We can either remember this as a formula or use the chain rule to figure it out.

Hence, the derivative of the function four sin six π₯ with respect to π₯ is four times six times cos six π₯, which simplifies to 24 cos six π₯. Recall that, for all constants π, the derivative of the function ππ₯ with respect to π₯ is just the constant π. Hence, the derivative of the function two π₯ with respect to π₯ is the constant two. Therefore, using LβHΓ΄pitalβs rule, the limit as π₯ approaches zero of four sin six π₯ over two π₯ is equal to the limit as π₯ approaches zero of 24 cos six π₯ over two.

Now if we use direct substitution, we obtain that the limit in question is equal to 24 cos of zero over two. Since cos of zero equals one, this simplifies to 24 over two, which equals 12. So the limit as π₯ approaches zero of four sin six π₯ over two π₯ is equal to 12.