Question Video: Finding an Unknown in a Piecewise-Defined Function Involving Trigonometric Ratios That Makes It Have a Limit at a Point Mathematics • Higher Education

Find all the values of π‘š for which the limit of the function exists as π‘₯ β†’ 0, when 𝑓(π‘₯) = βˆ’6π‘š βˆ’ 18 cos π‘₯, π‘₯ < 0 and 𝑓(π‘₯) = (tan 12π‘₯)/π‘šπ‘₯, π‘₯ > 0.

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Video Transcript

Find all the values of π‘š for which the limit of the function exists as π‘₯ approaches zero when 𝑓 of π‘₯ is equal to negative six π‘š minus 18 times the cos of π‘₯, when π‘₯ is less than zero, and 𝑓 of π‘₯ is equal to the tangent of 12π‘₯ divided by π‘š times π‘₯, when π‘₯ is greater than zero.

In this question, we’re given a piecewise-defined function 𝑓 of π‘₯, which contains an unknown value of π‘š. We need to determine all of the possible values of π‘š for which the limit of this function exists as π‘₯ is approaching zero. To answer this question, we should start by recalling what it means for the limit of a function at a point to exist. However, there’s many different ways of writing this and there’s many different ways for a limit to not exist. So instead, it’s usually a good idea to take a look at the function’s outputs around this value of π‘₯. This can give a useful idea about what might happen to the function.

And we can notice something very interesting. π‘₯ is equal to zero is one of the endpoints of the subdomains of this function. This means our function has a different definition when π‘₯ is less than zero and a different definition when π‘₯ is greater than zero. So for this limit to exist, we’re going to need to analyze its left and right limits. We can then recall that we say the limit as π‘₯ approaches π‘Ž of a function 𝑓 of π‘₯ exists and is equal to a finite value of 𝑙 if the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ both exist and are both equal to 𝑙.

This means we can check that the limit of a function exists by checking four things. First, we need to check that the limit from the right exists. Second, we need to check that this is equal to 𝑙. Third, we need to check that the limit from the left exists. And finally, we need to check that this is also equal to 𝑙.

To apply this to our function, we’re going to need to evaluate its left and right limits. Let’s start with the left limit. Since our value of π‘Ž is zero, this is the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯. And since we’re taking the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯, our values of π‘₯ are less than zero. So our function 𝑓 of π‘₯ is exactly equal to negative six π‘š minus 18 cos of π‘₯. And if the functions are equal, then their limits as π‘₯ approaches zero from the left will be equal. We just need to find the limit as π‘₯ approaches zero from the left of negative six π‘š minus 18 cos of π‘₯. And this is a constant plus a trigonometric function, so we can just do this by using direct substitution.

Substituting π‘₯ is equal to zero into this function, we get negative six π‘š minus 18 multiplied by the cos of zero. And the cos of zero is one, so this is just equal to negative six π‘š minus 18. And it’s worth noting this is a finite value for any real value of π‘š. So we’ve shown that this limit exists and we found our value of 𝑙. So let’s keep this in mind while we try to evaluate the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯. This time, since we’re taking the limit as π‘₯ approaches zero from the right, our values of π‘₯ are going to be greater than zero, which means our function 𝑓 of π‘₯ is exactly equal to the tan of 12π‘₯ divided by π‘š times π‘₯. And since the functions are equal to the right of π‘₯, the limit as π‘₯ approaches zero from the right will also be equal.

This means we just need to determine the limit as π‘₯ approaches zero from the right of the tan of 12π‘₯ divided by π‘š times π‘₯. And this limit is incredibly similar to one of our trigonometric limit results. We know for any real constant π‘Ž, the limit as π‘₯ approaches zero of the tan of π‘Žπ‘₯ divided by π‘₯ is equal to π‘Ž. And it’s worth noting these results will hold true as left and right limits since the left limit and the right limit need to be equal for the standard limit to exist. We can’t directly apply this, however, since we’re dividing by π‘š times π‘₯ in our denominator.

So we need to take a factor of one over π‘š outside of our limit. This then gives us one over π‘š multiplied by the limit as π‘₯ approaches zero from the right of the tan of 12π‘₯ divided by π‘₯. And by our limit result this is equal to the coefficient of π‘₯, which is 12. So we get 12 times one over π‘š, which is 12 over π‘š. Therefore, the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ exists, provided π‘š is not equal to zero. And we already knew that π‘š was nonzero since we’re dividing by π‘š in our definition of 𝑓 of π‘₯.

And this means we have one final thing we need to check. We need the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ to be equal to the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ for this limit to exist. We can substitute the expressions we found for each of these limits. We need negative six π‘š minus 18 to be equal to 12 divided by π‘š. To solve this equation, we’ll start by multiplying both sides of the equation through by π‘š and dividing through by six. This gives us that negative π‘š squared minus three π‘š is equal to two. We can then rearrange this quadratic to get π‘š squared plus three π‘š plus two is equal to zero and then factor the quadratic fully. We get π‘š plus one multiplied by π‘š plus two is equal to zero.

And of course, for a product to be equal to zero, one of the factors must be zero. So we either have π‘š is negative one or π‘š is negative two. And we can write these as a set, the set containing negative one and negative two, which is our final answer. Therefore, we were able to show there are only two values for which the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to negative six π‘š minus 18 cos of π‘₯ when π‘₯ is less than zero and 𝑓 of is equal to tan of 12π‘₯ divided by π‘šπ‘₯ when π‘₯ is greater than zero exists and must be in the set containing negative one and negative two.

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