00:01
Okay, so we want to find a slope of our tangent line given our x value.
00:05
So this time we're given that our x value is equal to 1.
00:08
So we know that this a here is 1.
00:11
So to find our slope, let's use this equation here.
00:15
So we have f of x and we need to find f of a.
00:18
We said that our a value is equal to 1.
00:20
So we have 3 times the square root of 1.
00:23
So that's equal to 3.
00:25
And then we have x minus a, that's equal to x minus 1.
00:29
Okay, so let's plug that into our equation.
00:31
Here.
00:32
So we have the limits as x approach is 1 of f of x that's minus 3, the square roots of x, minus f of a that's 3, all over x minus 1.
00:43
Okay? so i'm going to factor out that negative 3 that we have in common in our numerator.
00:51
So we're going to have a minus 3 and the square roots of x plus 1 over x minus 1.
00:57
So if you try to use the x up here, we're going to have division by 0.
01:01
So instead i'm going to multiply and divide by our conjugates in all.
01:04
And our numerator and our denominator.
01:10
Okay, so this is going to give us the limit as x approaches 1 of negative 3.
01:15
In our numerator, we have the difference of two squares.
01:18
That is, we're gonna rewrite this as the square root of x squared minus 1 squared.
01:23
So this is going to give us a x minus 1.
01:27
Okay, so that's times x minus 1, and this is all over x minus 1 times our conjugate.
01:38
Okay, so now we can cancel out our light terms, but is that going to help? this is going to give us a square root of 1 minus 1.
01:56
So we're going to have division by 0 once again...