00:01
Okay, so let's get started with the first function.
00:04
F of x equals 2x squared minus 5x plus 3 divided by x minus 1.
00:13
Now let's notice that the numerator here is divisible by x minus 1.
00:21
How can we write 2x squared minus 5x plus 3 as x minus 1 multiplied by something else? well, to do this, we are going to compute the roots of this guy.
00:36
So let's solve the quadratic equation to x squared minus 5x plus 3 equals 0.
00:43
Now here, thanks to the quadratic formula, we are going to have x equals 5 plus or minus 25 minus 4 multiplied by 6.
01:01
Over four.
01:03
So here we're going to have five minus one over four which is indeed one, a root of this guy.
01:12
And five, my and five plus one over four, which is three halves.
01:19
Perfect.
01:20
So this being shows that the numerator here can be written as two multiplied by x minus one multiplied by x minus one multiplied by x minus three halves in particular this pink shows what well this pink shows that limit as x goes to one of f of x is given by two multiplied by okay so here we are going to have two multiplied by x minus three halves evaluated at x equals one so 2 multiplied by 1 minus 3 halfs and this guy is negative 1...