Video: Finding the One-Sided Limits of a Function

Determine lim_(π‘₯ β†’ βˆ’9⁻) 𝑓(π‘₯) and lim_(π‘₯ β†’ βˆ’9⁺) 𝑓(π‘₯) given 𝑓(π‘₯) = {π‘₯ + 9, if π‘₯ ≀ βˆ’9 and 1/(π‘₯ + 9), if π‘₯ > βˆ’9.

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Video Transcript

Determine the left-sided limit as π‘₯ approaches negative nine of 𝑓 of π‘₯ and the right-sided limit as π‘₯ approaches negative nine of 𝑓 of π‘₯ given that 𝑓 of π‘₯ is equal to π‘₯ plus nine if π‘₯ is less than or equal to negative nine and one over π‘₯ plus nine if π‘₯ is greater than negative nine.

Here, we’ve been given a function defined piecewise over two intervals. For the left-sided limit, we’re approaching π‘₯ equals negative nine from the negative direction. Hence, π‘₯ is less than negative nine. For the right-sided limit, we’re approaching π‘₯ equals negative nine from the positive direction. And hence, π‘₯ is greater than negative nine. Since π‘₯ equals negative nine is the point between the two intervals of our piecewise function, for our left-sided limit will be in the first interval and for our right-sided limit will be in the second interval. Let’s work on finding the left-sided limit.

In this case, our function 𝑓 of π‘₯ is π‘₯ plus nine. We can find this limit by direct substitution of π‘₯ equals negative nine into our function. Doing so, we find that our answer is negative nine plus nine which is equal to zero. The left-sided limit as π‘₯ approaches negative nine of 𝑓 of π‘₯ is therefore zero. Now for the right-sided limit, here our function 𝑓 of π‘₯ is one over π‘₯ plus nine. Again, we tried direct substitution of π‘₯ equals negative nine into our function. This time, doing so gives us an answer of one over zero. And as we know, dividing one by zero cannot be evaluated to a numerical value. In cases such as this, we say that the limit does not exist. And so in a strict sense, this is the answer to our question.

To more fully understand our result, however, let’s look at a graphical plot of our function. Here, we have sketched our graph. And we know that, in the interval where π‘₯ is less than or equal to negative nine, we have a well behaved function. And we know that due to the solid dot here at negative nine π‘₯ is indeed defined at this point. For the other interval, we know that as π‘₯ approaches negative nine, we have a vertical asymptote. This means that the values of 𝑓 of π‘₯ get arbitrarily large. And this is often represented as infinity. In this sense, it is common to write that the right-sided limit as π‘₯ approaches negative nine of 𝑓 of π‘₯ is equal to positive infinity.

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