00:01
Okay, so first let's find a limit.
00:02
And then, so the limit of 2 over 3 times x plus 9, as x approaches 1 is equals 2 over 3 times 1 plus 9, which equals 29 over 3.
00:30
Okay? so next we want to use the epsilon delta definition to prove that limit.
00:48
Actually equals 29 over 3, okay? so what we want to prove is we want to prove that for each, for each imsyneal greater than 0, there exists a delta greater than 0, such that 2 over 3 times x plus 9 minus 29 over 3 is less than delta, whenever x minus 1, absolute value of x minus 1, is greater than 0, at less than ypsino.
01:58
So let's look at this inequality.
02:01
So we have 2 over 3 times x plus 9 minus 29 over 3, take the absolute value, which equals 2 over 3 times x plus 2 over 3.
02:17
Sorry, minus 2 over 3.
02:32
Okay, so we can factor out 2 over 3.
02:39
So 2 over 3, factor out times absolute value of x minus 1.
02:45
Okay, so for a given ipsenone that is greater 0, we can choose delta equals 2 over 3 times ipsyndo, okay? this works because x minus 1, absolute value of x minus 1, it's going to than 0, but less than delta, which equals 2 thirds times if you say no, implies 2 over 3 times x plus 9 minus 29 over 3 is greater than 0.
04:57
Last then, okay, so this, we can simplify this as 2 times 3 times, 2 over 3 times absolute value of x minus 1...