# Question Video: Discussing the Existence of the Limit of a Piecewise-Defined Function Containing Absolute Value in Its Rules at a Certain Point Mathematics • Higher Education

Given that π(π₯) = 1 + ((π₯Β² + 3π₯)/|π₯ + 3|), if β3 < π₯ < 0 and π(π₯) = 4π₯ + 2, if 0 < π₯ < 5, find lim_(π₯ β β3) π(π₯).

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### Video Transcript

Given that π of π₯ is equal to one plus π₯ squared plus three π₯ divided by the absolute value of π₯ plus three if π₯ is greater than negative three and less than zero and π of π₯ is equal to four π₯ plus two if π₯ is greater than zero and less than five, find the limit as π₯ approaches negative three of π of π₯.

In this question, weβre given a piecewise-defined function π of π₯. And weβre asked to find the limit as π₯ approaches negative three of this piecewise function π of π₯. And to do this, we can start by recalling we say the limit as π₯ approaches negative three of π of π₯ is equal to some finite value of π if the outputs of our function π of π₯ approach π as the input values of π₯ approach negative three from either side. And since weβre interested in the outputs of our function around the value of π₯ is equal to negative three, itβs always a good idea to see what happens to our function around this value of negative three.

And since π of π₯ is a piecewise-defined function, to do this, weβre going to need to look at the subdomains of our function. In particular, we can notice that negative three is one of the endpoints of the subdomains. And this can mean one of two things. Either our function π of π₯ changes definition to the left and to the right of negative three, or our function isnβt even defined one side of negative three. And for the piecewise-defined function π of π₯ given to us in the question, weβre in the latter version of this case. We can see this directly from the definition of π of π₯. However, it might be useful to find the domain of this function directly.

And we can do this by recalling the domain of a piecewise-defined function is the union of its subdomains. So we need to find the union of the open interval from negative three to zero with the open interval from zero to five. And we can calculate this directly. Itβs the open interval from negative three to five, where we remove the element zero.

And this then allows us to answer the question directly. We canβt take input values of π₯ less than negative three since these are not in the domain of our function π of π₯. So the limit as π₯ approaches negative three from the left of π of π₯ does not exist, which means the limit as π₯ approaches negative three of π of π₯ does not exist. In other words, negative three is a lower bound on the domain of π of π₯. And although this is enough to answer this question, itβs sometimes useful to see the property weβre directly using.

In this case, weβre using a property that we say the limit as π₯ approaches π of a function π of π₯ exists and is equal to a finite value of πΏ if both the limit as π₯ approaches π from the left of π of π₯ and the limit as π₯ approaches π from the right of π of π₯ both exist and are both equal to πΏ. And in this case, weβve shown we canβt take the limit as π₯ approaches negative three from the left of π of π₯, which means it does not exist.

Therefore, we were able to show if π of π₯ is equal to one plus π₯ squared plus three π₯ divided by the absolute value of π₯ plus three if π₯ is greater than negative three and less than zero and π of π₯ is equal to four π₯ plus two if π₯ is greater than zero and less than five, then the limit as π₯ approaches negative three of π of π₯ does not exist.