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Current time:0:00Total duration:3:46

Limit expression for the derivative of function (graphical)

Video transcript

with the graph of the function f as an aid evaluate the following limits so the first one is the limit as X approaches 3 of f of X minus F of 3 over X minus 3 so let's think about X minus x equals 3 is right over here this right over here is f of 3 or I say we could say F of 3 is 1 right over here that's the point 3 comma F of 3 and they're essentially trying to find the slope between an arbitrary X and that point as that X gets closer and closer to 3 so we can imagine an X that is above 3 that is say right over here well if we're trying to find the slope between this X comma f of X and 3 comma F of 3 we see that it gets this exact same form your endpoint is f of X so it's f of X minus F of 3 is your change in the vertical axis that's this distance right over here and you and we would divide by your change in the horizontal axis which is your change in X and that's going to be X minus 3 so that's the exact expression that we have up here when I picked this as an arbitrary X and we see that that slope just by looking at the line between those two intervals seems to be negative 2 and the slope was the same thing if we go on the other side if X was less than 3 then we also would have a slope of negative 2 either way we have a slope of negative 2 and that's important because this limit is just the limit as X approaches 3 so it can be as X approaches 3 from the from the positive direction or from the negative direction or from the negative direction but in either case the slope the slope as we get closer and closer to this point right over here is negative 2 negative 2 now let's think about what they're asking us here so we have a f of 8 so let's think we have 8 this is 8 comma f of 8 so that's 8 comma F of 8 right over there and they have f of 8 plus h so our temptation might be say hey 8 plus H is going to be someplace out here it's going to be something larger than 8 but notice they have the limit as H approaches 0 from the negative direction so approaching 0 from the negative direction means you're coming to 0 from below you're at negative 1 negative 0.5 negative 0.1 negative point 0 0 0 1 so H is actually going to be a negative number so 8 plus H would actually be we can just pick an arbitrary point it could be something like this right over here so this might be the value of 8 plus h and this would be the value of f of 8 plus h f of h 8 plus h so once again they're finding they're finding or this expression is the slope between these two points and then we are taking the limit as h approaches 0 from the negative direction so as H gets closer and closer to zero this down here moves further and further to the right and these points move closer and closer and closer together so this is really just an expression of the slope of the line roughly and we see that it's constant so what's the slope over the line over this interval well you can just eyeball it and see well look every time X changes by one our f of X changes by one so the slope of the line there is 1 would have been a completely different thing if this said limit as H approaches 0 from the positive direction then we would be then we would be looking at points over here and we would see that we would slowly approach it essentially a vertical slope kind of an infinite slope