Question Video: Finding the Sum of One-Sided Limits of a Function from Its Graph

Determine lim_(π‘₯ β†’ βˆ’5) 𝑓(π‘₯) + lim_(π‘₯ β†’ βˆ’7⁺) 𝑓(π‘₯) + lim_(π‘₯ β†’ 0⁻) 𝑓(π‘₯).

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Video Transcript

Determine the sum of the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five, the limit of 𝑓 of π‘₯ as π‘₯ approaches negative seven from the right, and the limit of 𝑓 of π‘₯ as π‘₯ approaches zero from the left.

We are given the graph of the function 𝑓 of π‘₯. We will use this graph to find the three limits needed to evaluate the sum. From the graph, we can see that 𝑓 is a piecewise defined function. It is defined by two subfunctions. One corresponding to the interval of π‘₯ values greater than or equal to negative seven but strictly less than zero and one corresponding to the interval of π‘₯ values strictly greater than zero but less than or equal to two. We are not given the algebraic expressions for these subfunctions, but just their graphs. Note that the small hollow circles on the 𝑦-axis at the points 𝑦 equals three and 𝑦 equals negative four indicate that the function 𝑓 is not defined at the π‘₯-value zero.

Let’s start by evaluating the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five. We know that if the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five exists, then it must be equal to the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five from the left and the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five from the right. From the graph of the subfunction corresponding to the values of π‘₯ greater than or equal to negative seven, but less than zero, we see that as π‘₯ approaches negative five from the left, i.e., takes values near negative five, but strictly less than negative five. And as π‘₯ approaches negative five from the right, i.e., takes values near negative five, but strictly greater than negative five. The function 𝑓 of π‘₯ approaches the value negative five. So the left and right limits of the function 𝑓 of π‘₯ as π‘₯ approaches negative five are both equal to negative five. So the limit of 𝑓 of π‘₯ as π‘₯ approaches negative five is also equal to negative five.

Now, let’s evaluate the limit of 𝑓 of π‘₯ as π‘₯ approaches negative seven from the right. From the graph of the subfunction corresponding to the values of π‘₯ greater than or equal to negative seven but less than zero, we see that as π‘₯ approaches negative seven from the right, i.e., takes values near negative seven, but strictly greater than negative seven. The function 𝑓 of π‘₯ approaches the value negative one. So the limit of 𝑓 of π‘₯ as π‘₯ approaches negative seven from the right is equal to negative one. Finally, let’s evaluate the limit of 𝑓 of π‘₯ as π‘₯ approaches zero from the left. Note that π‘₯ approaches zero from the left means that we are considering values of π‘₯ near zero but strictly less than zero. Therefore, we will consider the subfunction corresponding to the values of π‘₯ greater than or equal to negative seven but strictly less than zero when evaluating this limit.

As π‘₯ approaches zero from the left, i.e., takes the values near zero but strictly less than zero, the function 𝑓 of π‘₯ approaches the value negative four. So the limit of 𝑓 of π‘₯ as π‘₯ approaches zero from the left is equal to negative four. At this point, it might be worth noting that as π‘₯ approaches zero from the right, i.e., takes the values near zero but strictly greater than zero, the function 𝑓 of π‘₯ approaches the value three. So the limit of 𝑓 of π‘₯ as π‘₯ approaches zero from the right is equal to three. However, this is not needed to answer the question. In order to answer the question, we need to add together the values negative five, negative one, and negative four. Doing so, we obtain the value negative 10, which is our final answer.

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