### Video Transcript

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

In order to find a limit, our first instinct might be to use direct substitution. Letβs see how we could go about doing this in this question. Recall that, for all constants π and π, their functions π sin ππ₯ and π cos ππ₯ are continuous on the set of all real numbers. Therefore, both the functions three sin four π₯ in the numerator and two cos π₯ in the denominator of the limit weβre asked to find are continuous on the set of all real numbers. In particular, the functions three sin four π₯ and two cos π₯ are continuous at π₯ equals zero. Therefore, their quotient three sin four π₯ over two cos π₯ is continuous at π₯ equals zero, given that two cos of zero is not equal to zero.

We have that cos of zero equals one. Therefore, two cos of zero equals two, which is not equal to zero. Therefore, the quotient three sin four π₯ over two cos π₯ is continuous at π₯ equals zero.

Now recall that π of π₯ is continuous at π₯ equals π implies that the limit of π of π₯ as π₯ approaches π equals π evaluated at π. Therefore, we can use direct substitution to find the limit given to us in the question. Doing so, we obtain that the limit as π₯ approaches zero of three sin four π₯ over two cos π₯ is equal to three sin of four times zero over two cos of zero. Which simplifies to three sin of zero over two cos of zero.

Using the fact that sin of zero equals zero and cos of zero equals one, this is equal to three times zero over two times one. Which simplifies to zero over two, which is just zero. So we have found that the limit as π₯ approaches zero of three sin four π₯ over two cos π₯ is equal to zero.