Question Video: Finding the One-Sided Limit of a Piecewise-Defined Function Involving Trigonometric Ratios Mathematics

Video Transcript

Find the limit as 𝑥 approaches negative 𝜋 by six to the right of the function 𝑓 of 𝑥 given that 𝑓 of 𝑥 is equal to seven 𝑥 plus three sin 𝑥 all divided by the sin of five 𝑥 if 𝑥 is greater than negative 𝜋 by two and 𝑥 is less than zero, and 𝑓 of 𝑥 is equal to negative five cos of two 𝑥 plus seven if 𝑥 is greater than zero and 𝑥 is less than 𝜋 by two.

The question wants us to calculate the right-hand limit of a piecewise-defined function, 𝑓 of 𝑥. Since the limit we’re asked to evaluate has 𝑥 approaching negative 𝜋 by six to the right, our values of 𝑥 will always be bigger than negative 𝜋 over six. And since the limit we want to evaluate has 𝑥 approaching negative 𝜋 over six from the right, we want our values of 𝑥 to get closer and closer to negative 𝜋 over six. This means our values of 𝑥 will eventually be negative when they get closer and closer to negative 𝜋 over six. So, when we’re evaluating this limit, we know that 𝑥 is always greater than negative 𝜋 over six, but it’s eventually less than zero.

And we can see from our piecewise definition of the function 𝑓 of 𝑥, if 𝑥 is greater than negative 𝜋 over two and less than zero, our function 𝑓 of 𝑥 is equal to seven 𝑥 plus three sin 𝑥 all divided by the sin of five 𝑥. This means just to the right of our value of negative 𝜋 over six, our function 𝑓 of 𝑥 is exactly equal to seven 𝑥 plus three sin 𝑥 all divided by the sin of five 𝑥. And if these functions are equal just to the right of negative 𝜋 over six, then their limits as 𝑥 approaches negative 𝜋 over six from the right will also be equal. We’re now asked to calculate the limit of a combination of polynomial and standard trigonometric functions. We can do this by using direct substitution.

Substituting 𝑥 is equal to negative 𝜋 over six gives us seven times negative 𝜋 over six plus three times the sin of negative 𝜋 over six all divided by the sin of five times negative 𝜋 over six. We’re now ready to evaluate this expression. We know that the sine is an odd function, so the sin of negative 𝜋 over six is equal to negative the sin of 𝜋 over six. And the sin of 𝜋 over six is a standard result which we should know; it’s one-half. So, the sin of negative 𝜋 over six is negative a half. We can do the same to calculate the sin of negative five 𝜋 by six. It’s equal to negative the sin of five 𝜋 by six. And we know the sin of five 𝜋 by six is equal to a half. So, our denominator evaluates to give us negative a half.

This gives us negative seven 𝜋 over six minus three over two all divided by negative a half. And instead of dividing by negative half, we can multiply by the reciprocal. Multiplying by the reciprocal gives us negative seven 𝜋 by six minus three over two all multiplied by negative two. And then, we distribute over the parentheses and reorder our terms to get three plus seven 𝜋 by three. Therefore, we’ve shown the limit as 𝑥 approaches negative 𝜋 over six from the right of our function 𝑓 of 𝑥. Where 𝑓 of 𝑥 is equal to seven 𝑥 plus three sin 𝑥 over the sin of five 𝑥 if 𝑥 is greater than negative 𝜋 over two and 𝑥 is less than zero. And 𝑓 of 𝑥 is equal to negative five cos of two 𝑥 plus seven if 𝑥 is greater than zero and 𝑥 is less than 𝜋 over two. Is equal to three plus seven 𝜋 by three.