Question Video: Finding the Limit of a Piecewise-Defined Function Containing Absolute Value in One of Its Rules at a Certain Point | Nagwa Question Video: Finding the Limit of a Piecewise-Defined Function Containing Absolute Value in One of Its Rules at a Certain Point | Nagwa

Question Video: Finding the Limit of a Piecewise-Defined Function Containing Absolute Value in One of Its Rules at a Certain Point Mathematics • Second Year of Secondary School

Find lim_(𝑥 → 4) 𝑓(𝑥) if 𝑓(𝑥) = 5 + ((𝑥² + 3𝑥)/|𝑥 + 3|), if −4 < 𝑥 < 0 and 𝑓(𝑥) = 5𝑥 + 4, if 0 < 𝑥 < 4.

02:20

Video Transcript

Find the limit as 𝑥 approaches four of 𝑓 of 𝑥 if 𝑓 of 𝑥 is equal to five plus 𝑥 squared plus three 𝑥 over the absolute value of 𝑥 plus three if 𝑥 is greater than negative four and less than zero and 𝑓 of 𝑥 is equal to five 𝑥 plus four if 𝑥 is greater than zero and less than four.

In this question, we’re given a piecewise-defined function 𝑓 of 𝑥 and asked to determine the limit as 𝑥 approaches four of 𝑓 of 𝑥. And to answer this question, we need to start by noticing one thing. We’re asked to find the limit as 𝑥 approaches four of 𝑓 of 𝑥, and four is one of the endpoints of the subdomains of our piecewise function 𝑓 of 𝑥. And since this is the endpoint of one of the subdomains, this means our function 𝑓 of 𝑥 will change definition to the left and to the right of four. So we need to consider the left and right limits of 𝑓 of 𝑥 as 𝑥 approaches four.

We can do this by recalling the following property. We say the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 is equal to a finite value of 𝐿 if the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 both exist and are both equal to 𝐿. And at this point, we can notice something interesting. We can find the limit as 𝑥 approaches four from the left of 𝑓 of 𝑥, since our function 𝑓 of 𝑥 is equal to the linear function five 𝑥 plus four when 𝑥 is between zero and four. So when we take the limit as 𝑥 approaches four from the left of 𝑓 of 𝑥, we can choose our values of 𝑥 to be closer and closer to four. So we can choose them between zero and four. This means its limit is the same as the limit as 𝑥 approaches four from the left of five 𝑥 plus four.

And we can then evaluate this limit by direct substitution. We substitute 𝑥 is equal to four into the near function. We get five times four plus four, which is equal to 24. So the left limit exists and is equal to 24. However, none of this was necessary. We can notice directly from the definition of 𝑓 of 𝑥 that the limit as 𝑥 approaches four from the right of 𝑓 of 𝑥 does not exist. And this is a direct consequence from the fact no input value of 𝑥 greater than four is in the domain of 𝑓 of 𝑥.

And it might be useful to just find the domain of 𝑓 to be safe. We can do this by finding the union of its subdomains. This gives us the open interval from negative four to four, where we remove the point zero. And we can see that no value greater than four is in this set. So we can’t take the limit as 𝑥 approaches four from the right of 𝑓 of 𝑥.

Therefore, we can say that the limit does not exist because the limit as 𝑥 approaches four from the right of 𝑓 of 𝑥 also doesn’t exist.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

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