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

Find lim_(π₯ β β(πβΊ/6)) π(π₯) given π(π₯) = (7π₯ + 3 sin π₯)/(sin 5π₯) if β(π/2) < π₯ < 0, and π(π₯) = β5 cos 2π₯ + 7 if 0 < π₯ < (π/2).

03:05

### Video Transcript

Find the limit as π₯ approaches negative π by six to the right of the function π of π₯ given that π 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 by two and π₯ is less than zero, and π 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 by two.

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

And we can see from our piecewise definition of the function π of π₯, if π₯ is greater than negative π over two and less than zero, our function π 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 over six, our function π 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 over six, then their limits as π₯ approaches negative π over six from the right will also be equal. Weβ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 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. Weβ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; itβ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. Itβ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, weβve shown the limit as π₯ approaches negative π over six from the right of our function π of π₯. Where π of π₯ is equal to seven π₯ plus three sin π₯ over the sin of five π₯ if π₯ is greater than negative π over two and π₯ is less than zero. And π 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 over two. Is equal to three plus seven π by three.