Video: Discussing the Existence of One-Sided Limits of Piecewise Defined Functions

Discuss the existence of lim_(π‘₯ β†’ 4⁻) 𝑓(π‘₯) given 𝑓(π‘₯) = |π‘₯ βˆ’ 1| + 2 if βˆ’1 < π‘₯ < 2 and 𝑓(π‘₯) = (π‘₯Β² + 2π‘₯ βˆ’ 8)/(π‘₯Β² βˆ’ 2π‘₯) if 2 < π‘₯ < 4.

03:47

Video Transcript

Discuss the existence of the limit as π‘₯ approaches four from the left of 𝑓 of π‘₯ given that 𝑓 of π‘₯ is equal to the absolute value of π‘₯ minus one plus two if π‘₯ is greater than negative one and π‘₯ is less than two. And 𝑓 of π‘₯ is equal to π‘₯ squared plus two π‘₯ minus eight all divided by π‘₯ squared minus two π‘₯ if π‘₯ is greater than two and π‘₯ is less than four.

The question wants us to discuss the existence of the limit as π‘₯ approaches four from the left of our piecewise-defined function 𝑓 of π‘₯. And we remember, we say that this limit exists if it gets closer and closer to some finite value 𝐿. In particular, if it’s unbounded or it does not get closer and closer to some finite value of 𝐿, then we’ll say that the limit does not exist. One way we can check the existence of a limit is to try and evaluate our limit. If we can evaluate our limit and find that it’s equal to some finite value of 𝐿, we’ve shown that the limit exists. We want to find the limit of our function 𝑓 of π‘₯ as π‘₯ approaches four from the left.

Since π‘₯ is approaching four from the left, we must have that π‘₯ is less than four. And since we’re taking the limit as π‘₯ approaches four, we want π‘₯ to get closer and closer to four. So eventually, our values of π‘₯ will be bigger than two. So, when we’re taking the limit as π‘₯ approaches four from the left of our function 𝑓 of π‘₯, our values of π‘₯ will always be less than four, and eventually they’ll be bigger than two. And we see from our piecewise definition of the function 𝑓 of π‘₯, if π‘₯ is greater than two and less than four, our function 𝑓 of π‘₯ is exactly equal to the rational function π‘₯ squared plus two π‘₯ minus eight all divided by π‘₯ squared minus two π‘₯. So, both of these functions are equal as π‘₯ approaches four from the left. That means their limits as π‘₯ approaches four from the left will also be equal.

So, we now need to evaluate the limit of our rational function. We can try using direct substitution. Substituting π‘₯ is equal to four gives us four squared plus two times four minus eight all divided by four squared minus two times four. Evaluating and simplifying this expression gives us that our limit was equal to two. So, we’ve shown the limit as π‘₯ approaches four from the left of 𝑓 of π‘₯ is equal to two. This is a finite value, so our limit exists. Therefore, we’ve shown the limit as π‘₯ approaches four from the left of 𝑓 of π‘₯ exists and is equal to two.

To help us see why this limit is equal to two, let’s sketch a graph of our function 𝑓 of π‘₯. We’ll start with the graph of 𝑦 is equal to the absolute value of π‘₯ minus one plus two, where π‘₯ is between negative one and two. We can think of this graph as the graph of 𝑦 is equal to the absolute value of π‘₯ except we translate it one unit right and two units up. And our curve does not include the end points. We represent this with the hollow circles.

Next, we sketch a segment of the curve 𝑦 is equal to π‘₯ squared plus two π‘₯ minus eight divided by π‘₯ squared minus two π‘₯. We sketch the part where π‘₯ is greater than two and π‘₯ is less than four. By factoring the quadratic in our numerator and our denominator, we can simplify our expression. Cancelling the shared factor of π‘₯ minus two gives us π‘₯ plus four divided by π‘₯. Then, we can divide through by our denominator of π‘₯ to get one plus four divided by π‘₯.

And we could use this to sketch the graph of our rational function. It’s the same as the graph of one over π‘₯ except we stretch it and then we translate it up one unit. This means that our curve, 𝑦 is equal to our piecewise function 𝑓 of π‘₯, is these two curves combined. And we want to know what happens to the limit of π‘₯ as π‘₯ approaches four from the left. And we can see as π‘₯ is getting closer and closer to four from the left-hand side, our output is getting closer and closer to four. This gives us a graphical justification of the existence of our limit. In conclusion, we’ve shown the limit as π‘₯ approaches four from the left of our function 𝑓 of π‘₯ exists and is equal to two.

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