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

Find lim_(π‘₯ β†’ πœ‹β») 𝑓(π‘₯) given 𝑓(π‘₯) = {(5π‘₯ cos 5π‘₯ +2 sin 5π‘₯)/π‘₯, if 0 < π‘₯ < πœ‹/2 and 4/(2 cos 9π‘₯ + πœ‹), if πœ‹/2 < π‘₯ < πœ‹.


Video Transcript

Find the left-sided limit as π‘₯ approaches πœ‹ of 𝑓 of π‘₯ given that 𝑓 of π‘₯ is equal to five π‘₯ cos five π‘₯ plus two sin five π‘₯ over π‘₯ if π‘₯ is greater than zero and less than πœ‹ over two and four over two cos nine π‘₯ plus πœ‹ if π‘₯ is greater than πœ‹ over two and less than πœ‹.

Here, we have a piecewise function 𝑓 of π‘₯ defined over two different intervals. Our function is clearly undefined when π‘₯ is less than or equal to zero or greater than or equal to πœ‹. We can also note that due to these strict inequality symbols 𝑓 of π‘₯ is also undefined when π‘₯ is equal to πœ‹ over two. Now, our question is asking us for the left-sided limit as we can see from this minus symbol. This means that we’re approaching a value of π‘₯ equals πœ‹ from the negative direction. And π‘₯ is strictly less than πœ‹. Even though we know that π‘₯ equals πœ‹ is not in the domain of our function, we can still try and find a limit since the limits concern values of π‘₯ which are arbitrarily close to πœ‹ but not equal to πœ‹.

Now, we know that the values of π‘₯ that we’re interested in are less than πœ‹ but very close to this value. Hence, the interval of our function that we’re interested in is this one where 𝑓 of π‘₯ is equal to four over two cos of nine π‘₯ plus πœ‹. And we can see this from our inequality. Since this is the interval for which π‘₯ is just less than πœ‹, we proceed to evaluate our limit as follows. We take a direct substitution of π‘₯ equals πœ‹ into our function. Looking at the term cos of nine πœ‹, we recall that cosine is a periodic function and repeats itself over a period of two πœ‹ radians. This means that cos of nine πœ‹ is equal to cause of πœ‹. And of course this is equal to negative one.

If we perform this substitution, we find that our answer becomes the following quotient. And we then arrive at an answer of four over negative two plus πœ‹. We have now answered the question. And we have found the left-sided limit as π‘₯ approaches πœ‹ of the function 𝑓 of π‘₯. In some cases, it might be difficult to sketch your function. And here we’ve demonstrated taking a one-sided limit without plotting our function graphically. It’s also worth reflecting back on the fact that since 𝑓 of π‘₯ is undefined when π‘₯ is greater than πœ‹ and indeed when π‘₯ is equal to πœ‹, then the right-sided limit as π‘₯ approaches πœ‹ of 𝑓 of π‘₯ does not exist. And indeed the normal limit as π‘₯ approaches πœ‹ of 𝑓 of π‘₯ cannot be said to exist either.

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