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

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

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Find lim_(π‘₯ β†’ βˆ’9) 𝑓(π‘₯), where 𝑓(π‘₯) = βˆ’8 + |π‘₯ + 9| if π‘₯ β‰  βˆ’9 and 𝑓(π‘₯) = βˆ’7 when π‘₯ = βˆ’9.

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Video Transcript

Find the limit as π‘₯ approaches negative nine of 𝑓 of π‘₯, where 𝑓 of π‘₯ is equal to negative eight plus the absolute value of π‘₯ plus nine if π‘₯ is not equal to negative nine and 𝑓 of π‘₯ is equal to negative seven when π‘₯ is equal to negative nine.

In this question, we’re given a piecewise-defined function 𝑓 of π‘₯ and we’re asked to find the limit as π‘₯ approaches negative nine of this function 𝑓 of π‘₯. And there are a few different methods we could use to evaluate this limit. We’ll only go through one of these.

Since our function 𝑓 of π‘₯ is a piecewise-defined function, we might be tempted to start by looking at the left and right limits of 𝑓 of π‘₯ as π‘₯ approaches negative nine. And this would work and give us the correct answer. However, we can make this easier by noticing something about the subdomains.

Our function 𝑓 of π‘₯ has two subdomains: π‘₯ not being equal to negative nine and π‘₯ being equal to negative nine. And remember, when we’re taking the limit of a function as π‘₯ approaches negative nine, we don’t need to know what happens to our function when π‘₯ is equal to negative nine. We’re only interested in the outputs of our function when π‘₯ gets closer and closer to negative nine from either direction. Therefore, our function 𝑓 of π‘₯ is equal to negative eight plus the absolute value of π‘₯ plus nine when π‘₯ is not equal to negative nine. So, their limits as π‘₯ approaches negative nine must also be equal.

And now we’re just evaluating the limit of a sum of a constant function and an absolute value function. And we recall we can evaluate the limit of absolute value functions by using direct substitution. So we substitute π‘₯ is equal to negative nine into our function. This gives us negative eight plus the absolute value of negative nine plus nine. And since negative nine plus nine is equal to zero and the absolute value of zero is just zero, this just gives us negative eight, which is our final answer.

Therefore, we were able to show the limit as π‘₯ approaches negative nine of 𝑓 of π‘₯ is equal to negative eight plus the absolute value of π‘₯ plus nine when π‘₯ is not equal to negative nine and 𝑓 of π‘₯ is equal to negative seven when π‘₯ is equal to negative nine is negative eight.

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