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Analyzing unbounded limits: rational function

Sal analyzes the behavior of f(x)=-1/(x-1)² around its asymptote at x=1.

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Video transcript

- [Voiceover] Let f of x be equal to negative one over x minus one squared. Select the correct description of the one-sided limits of f at x equals one. And so we can see, we have a bunch of choices where we're approaching x from the right-hand side and we're approaching x from the left-hand side. And we're trying to figure out do we get unbounded on either of those, in the positive, towards positive infinity or negative infinity. And there's a couple of ways to tackle it. The most straightforward, well, let's just consider each of these separately. So we can think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one, as x approaches one from the left-hand side. This is from the right-hand side. This is from the left-hand side. So I'm just gonna make a table and try out some values as we approach, as we approach one from the different sides, x, f of x, and I'll do the same thing over here. So, we are going to have our x and have our f of x and if we approach one from the right-hand side here, that would be approaching one from above, so we could try 1.1, we could try 1.01. Now f of 1.1 is negative one over 1.1 minus one squared. So see this denominator here is going to be .1 squared. So this is going to be, this is going to be 0.01, and so this is going to be negative 100. So let me just write that down. That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, then this denominator this is going to be, this is the same thing as 0.01 squared, which is the same thing as 0.0001, 1/10000. And so the negative one 1/10000 is going to be negative 10,000. So, let's just write that down, negative 10,000. And so this looks like, as we get closer, 'cause notice, as I'm going here I am approaching one from the positive direction, I'm getting closer and closer to one from above and I'm going unbounded towards negative infinity. So this looks like it is negative infinity. Now we can do the same thing from the left-hand side. I could do 0.9, I could do 0.99. Now 0.9 is actually also going to get me negative 100 'cause 0.9 minus one is going to be negative .1 but then when you square it the negative goes away so you get a .01 and then one divided by that is 100 but you have the negative, so this is also negative 100. And if you don't follow those calculations, I'll do it, let me do it one more time just so you see it clearly. This is going to be negative one over, so now I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below from the left-hand side. So this is going to be 0.99 minus one squared. Well, 0.99 minus one is, is going to be negative 1/100, so this is going to be negative 0.01 squared. When you square it the negative goes away and you're left with 1/10000. So this is going to be 0.0001 and so when you evaluate this you get 10,000. So that, or sorry, you get negative 10,000. So in either case, regardless of which direction we approach from, we are approaching negative infinity. So that is this choice right over here. Now there's other ways you could have tackled this if you just look at, kind of, the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive. Let's ignore this negative for the time being. That negative's out front. This numerator, this one is always going to be positive. Down here, we're taking at x equals one, while this becomes zero and the whole expression becomes undefined, but as we approach one, x minus one could be positive or negative as we see over here, but then when we square it, this is going to become positive as well. So the denominator is going to be positive for any x other than one. So positive divided by positive is gonna be positive but then we have a negative out front. So this thing is going to be negative for any x other than one, and it's actually not defined at x equals one. And so you could, from that, you could deduce, well, okay then, we can only go to negative infinity there's actually no way to get positive values for this function.