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Main content
Current time:0:00Total duration:4:09
AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
,
LIM‑1.D.2 (EK)

Video transcript

let's find the limit of f of X times H of X as X approaches zero all right and we have graphical depictions of the graphs y equals f of X and y equals H of X and we know from our limit properties that this is going to be the same thing as the limit as X approaches 0 of f of X x times the limit as X approaches 0 of H of X and let's think about what each of these are so let's first think about f of X right over here so on f of X is X approaches 0 notice the function itself isn't defined there but we see when we approach from the left we are approaching the the function seems to be approaching the value of negative 1 right over here and as we approach from the right the function seems to be approaching the value of of negative 1 so the limit here this limit here is negative 1 as we approach from the left we're approaching negative 1 superwatch from the right the value of the function seems to be approaching negative 1 now what about H of X well H of X we have down here as X approaches 0 as X approaches 0 the function is defined at x equals 0 it looks like it is equal to 1 and you could and the limit is also equal to 1 we can see that as we approach it from the left we are approaching 1 as we approach from the right we're approaching 1 as we approach x equals 0 from the left we approach the function approaches 1 as we approach x equals 0 from the right the function itself is approaching 1 and it makes sense that the function is defined there and the is defined at x equals 0 and the limit as X approaches 0 is equal to the same as the is equal to the value of the function at that point because this is a continuous function so this is this is 1 and so negative 1 times 1 is going to be equal to is equal to negative 1 so that is equal to negative 1 let's do one more all right so these are both looks like continuous functions so we have the limit as X approaches 0 of H of X over G of X so once again using our limit properties this is going to be the same thing as the limit of H of X as X approaches 0 over the limit of G of X as X approaches 0 now what's the limit of H of X as X approaches 0 this is let's see as we approach 0 from the left as we approach x equals 0 from the left our function seems to be approaching 4 and if you approach x equals 0 from the right our function seems to be approaching 4 and that's also what the value of the function is at x equals 0 that makes sense because this is a continuous function so the limit as we approach x equals 0 should be the same as the value of the function at x equals 0 so this top this is going to be 4 and let's think about the limit of G of X is x equals as X approaches 0 so from the left it looks like as X approaches 0 the value of the function is approaching 0 and as X approaches 0 from the right the value of the function is also approaching 0 which happens to also be which also happens to be G of 0 G of 0 is also 0 and that makes sense that the limit and the actual value of the function at that point is the same because it's continuous so this also is 0 but now we're in a strange situation we have to take 4 and divide it by 0 so this limit will not exist because we can't take 4 and divide it by 0 so even though the limit of H of X is x equals as X approaches 0 exist and the limit of G of X as X approaches 0 exists we can't divide 4 by 0 so this whole entire limit does not exist does does not exist and actually if you were to plot H of X over G of X if you were to plot that graph you would see it even clearer that that limit does not exist you would actually be able to see it graphically
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