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Analyzing unbounded limits: rational function | AP Calculus AB | Khan Academy


3m read
·Nov 11, 2024

Let f of x be equal to negative 1 over x minus 1 squared. Select the correct description of the one-sided limits of f at x equals 1.

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. We're trying to figure out do we get unbounded on either of those in the positive direction towards positive infinity or negative infinity.

There's a couple of ways to tackle it. The most straightforward, well, let's just consider each of these separately. We could think about the limit of f of x as x approaches 1 from the positive direction and the limit of f of x as x approaches 1 from the left-hand side.

This is from the right-hand side, this is from the left-hand side. So I'm just going to make a table and try out some values 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. If we approach 1 from the right-hand side here, that would be approaching 1 from above. So we could try 1.1, we could try 1.01.

Now, f of 1.1 is negative 1 over 1.1 minus 1 squared.

So see, this denominator here is going to be 0.1 squared, so 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. If x is 1.01, well, this is going to be negative 1 over 1.01 minus 1 squared.

Well, in this denominator, this is going to be 0.01 squared, which is the same thing as 0.0001, one ten-thousandth. And so negative 1 over one ten-thousandth is going to be negative ten thousand.

So let's just write that down: negative 10,000. This looks like, as we get closer—because notice as I'm going here, I am approaching 1 from the positive direction—I'm getting closer and closer to 1 from above and I'm going unbounded towards negative infinity.

So this looks like it is negative infinity. Now we could 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 because 0.9 minus 1 is going to be negative 0.1, but then when you square it, the negative goes away. So you get 0.01, and then 1 divided by that is 100, but you have the negative, so this is also negative 100.

If you don't follow those calculations, I'll do it. Let me do it one more time just so you see it clearly. There's going to be negative 1 over, so now I'm doing x is equal to 0.99.

So I'm getting even closer to 1, but I'm approaching from below, from the left-hand side. So this is going to be 0.99 minus 1 squared.

Well, 0.99 minus 1 is going to be negative 0.01 squared. Well, when you square it, the negative goes away, and you're left with one ten-thousandth.

So this is going to be 0.0001. And so when you evaluate this, 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 are 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 is out front. This numerator, this 1, is always going to be positive.

Down here, we're taking the limit as x equals 1. Well, this becomes 0, and the whole expression becomes undefined.

But as we approach 1, x minus 1 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 going to be positive, but then you 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, deduce, well okay then we can only go to negative infinity. There's actually no way to get positive values for this function.

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