If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Limit expression for the derivative of a linear function

## Video transcript

let G of X equal negative 4x plus 7 what is the value of the limit as X approaches negative 1 of all of this so before we think about this let's just visualize the line and then we can think about what they're asking here so let me draw some axes here so this is my vertical axis and this is my horizontal axis my horizontal axis and let's say that this is my x-axis we'll label out the x axis I'll graph G of X G of X is going to have a positive I guess we could say y-intercept or vertical axis intercept it's going to have a slope of negative 4 so it's going to look something like this let me draw my best so it's going to look something something like that and we already know that the slope the slope here is going to be negative 4 we get that right from this slope-intercept form of the equation slope is equal to negative 4 and they asked us what is the limit as X approaches negative 1 of all of this kind of stuff so let's plot the point negative 1 so when X is negative 1 so that's this point right over here and this point right over here would be the point negative 1 comma G of negative 1 and let me label everything else so I could call this my y-axis I could call this graph this is the graph of y is equal to G of X so what they're doing right over here is they're finding the slope between an arbitrary point X G of X and this point right over here so let's do that so let's take another X so let's say this is X this would be the point this would be the point X G of X and this expression right over here notice it is your change in the vertical axis your change in the vertical axis that would be your G of X let me make it this way so this would be your your change in the vertical axis that would be G of X minus G of negative 1 and then that's over actually let me write it this way so that you can keep track of the colors minus G of negative 1 all of that over your change in the horizontal axis well your change in the horizontal axis is this distance this distance which is the same thing as this distance notice your change in vertical over change in horizontal change in vertical over change in horizontal we're viewing the Green Point as the end point so it's going to be X minus negative one and these this is the exact same expression these are the exact same expression you could simplify the minus negative one and this becomes plus this could become a plus one but these are the exact same expression so this is the expression really for the slope between negative one and G of negative one and an arbitrary X well we already know that that no matter what X you you pick the slope between XG of X and this point right over here is going to be constant it's going to be the slope of the line it's going to be equal to negative four this thing is going to be equal to negative four it's going to be equal to negative four doesn't matter how close how close X gets and whether X comes from a bit from the right or there X comes from the left so this thing taking the limit of this this just gets you to negative four it's really just the slope of the line so even if you were to take the limit as X approaches negative one this X gets closer and closer and closer to negative one well then these points are just going to get closer and closer and closer but every time you calculate the slope it's just going to be the slope of the line which is negative four now you could also do this algebraically and let's try to do it algebraically so let's actually just take the limit the limit as X approaches negative one of G of X well they already told us what G of X is it is negative 4x plus seven minus G of negative one so that's minus what is G of negative one negative one times 4 is positive 4 positive 4 plus seven is 1111 all of that over all of that over X plus 1 all of that over X plus 1 and that's really that's where it's really X minus negative 1 is if you want to think of it that way but just write x plus one this way here so this is going to be equal to the limit as X approaches negative 1 of in our numerator let's see 7 minus 11 is negative 4 we can factor out a negative 4 it's negative 4 times X plus 1 all of that over X plus 1 and then since we're just trying to find the limit as X approaches negative 1 so we can we can cancel those out and what this is going to be this is going to be nonzero for any X for any x value other than negative 1 and so this is going to be equal to negative 4 so either way we get negative 4 but if you just realize hey this is a line it's going to have a constant slope this is just the slope of between some arbitrary point on the line and the point negative 1 comma 11 really or negative 1 comma 11 you say well that's just going to be the same as the slope of line it's negative 4