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

Find lim_(π‘₯ β†’ βˆ’(πœ‹βΊ/6)) 𝑓(π‘₯) given 𝑓(π‘₯) = (7π‘₯ + 3 sin π‘₯)/(sin 5π‘₯) if βˆ’(πœ‹/2) < π‘₯ < 0, and 𝑓(π‘₯) = βˆ’5 cos 2π‘₯ + 7 if 0 < π‘₯ < (πœ‹/2).

03:05

Video Transcript

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

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

And we can see from our piecewise definition of the function 𝑓 of π‘₯, if π‘₯ is greater than negative πœ‹ over two and less than zero, our function 𝑓 of π‘₯ is equal to seven π‘₯ plus three sin π‘₯ all divided by the sin of five π‘₯. This means just to the right of our value of negative πœ‹ over six, our function 𝑓 of π‘₯ is exactly equal to seven π‘₯ plus three sin π‘₯ all divided by the sin of five π‘₯. And if these functions are equal just to the right of negative πœ‹ over six, then their limits as π‘₯ approaches negative πœ‹ 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 π‘₯ is equal to negative πœ‹ over six gives us seven times negative πœ‹ over six plus three times the sin of negative πœ‹ over six all divided by the sin of five times negative πœ‹ 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œ‹ over six is equal to negative the sin of πœ‹ over six. And the sin of πœ‹ over six is a standard result which we should know; it’s one-half. So, the sin of negative πœ‹ over six is negative a half. We can do the same to calculate the sin of negative five πœ‹ by six. It’s equal to negative the sin of five πœ‹ by six. And we know the sin of five πœ‹ by six is equal to a half. So, our denominator evaluates to give us negative a half.

This gives us negative seven πœ‹ 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œ‹ 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œ‹ by three. Therefore, we’ve shown the limit as π‘₯ approaches negative πœ‹ over six from the right of our function 𝑓 of π‘₯. Where 𝑓 of π‘₯ is equal to seven π‘₯ plus three sin π‘₯ over the sin of five π‘₯ if π‘₯ is greater than negative πœ‹ over two and π‘₯ is less than zero. And 𝑓 of π‘₯ is equal to negative five cos of two π‘₯ plus seven if π‘₯ is greater than zero and π‘₯ is less than πœ‹ over two. Is equal to three plus seven πœ‹ by three.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.