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# 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. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of x. g of x is going to have a positive-- I guess you would 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 like that. And we already know 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 ask 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, g of negative 1. 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 x, g of x. And this expression right over here, notice it is your change in the vertical axis. That would be your g of x. Let me make it this way. So this would be 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 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, which is the same thing as this distance. Notice your change in vertical over change in horizontal, change in vertical over change in horizontal, reviewing the green point as the endpoint. So it's going to be x minus negative 1. And this is the exact same expression. These are the exact same expression. You can simplify them, minus negative 1, and this becomes plus. This could become a plus 1. But these are the exact same expression. So this is the expression, really, for the slope between negative 1 and g of negative 1 and an arbitrary x. Well, we already know that no matter what x you pick, the slope between x, g of , 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 4. This thing is going to be equal to negative 4. It's going to be equal to negative 4. Doesn't matter how close x gets, and weather x comes from the right or whether x comes from the left. So this thing, taking the limit of this, this just gets you to negative 4. It's really just the slope of the line. So even if you were to take the limit as x approaches negative 1, as x gets closer and closer and closer to negative 1, 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 4. Now, you could also do this algebraically. And let's try to do it algebraically. So let's actually just take the limit as x approaches negative 1 of g of x. Well, they already told us what g of x is. It is negative 4x plus 7, minus g of negative 1. So that's minus, what is g of negative 1? Negative 1 times 4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1, all of that over x plus 1. And that's really x minus negative 1, is you want to think of it that way. But I'll just write x plus 1 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 a 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 cancel those out. And this is going to be non-zero 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, 11, really. Negative 1, 11, you say, well that's just going to the same as the slope of a line. It's negative 4.