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Current time:0:00Total duration:5:03

AP.CALC:

LIM‑2 (EU)

, LIM‑2.D (LO)

, LIM‑2.D.1 (EK)

, LIM‑2.D.2 (EK)

let f of X is e 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 and we're trying to figure out do we get unbounded on either of those in the positive 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 could think about the limit of f of X as X approaches 1 from the positive direction and limit of f of X as X approaches 1 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 gonna make a table and try out some values as we approach as we approach 1 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 1 from the right-hand side here that would be approaching 1 from above so we could try one point 1 we could try one point zero one now F of one point one is negative one over one point one minus one squared so see this denominator here is going to be point one squared so this is going to be this is going to be zero point zero one and so this is going to be negative one hundred so let me just write that down that's going to be negative one hundred so if we if X is one point zero one well this is going to be negative one over one point zero one minus one squared well in this denominator this is going to be point this is the same thing as 0.01 squared which is the same thing is 0.0001 one ten-thousandth and so the negative one one ten-thousandth is going to be negative ten thousand so let's just write that down negative ten thousand and so this looks like is 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 one 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 point 1 but then when you square it the negative goes away so you get a point 0 1 and then 1 divided by that is 100 whether you have a 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 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 is going to be negative one hundredth so this is going to be negative 0.01 squared well when you square it the negative goes away and you're left with one 10,000 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 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 negatives out front this numerator this one is always going to be positive down here we're taking the at x equals one well 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 a 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 you could to do as 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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