00:01
Here we're given the switch on the form minus 1 power of n and now x minus 3 about n divided by 2 n plus 1 goes from 0 to infinity and the first time we need to use the ratio test here and then here we need to compute the limit of the am plus 1 depending by an when n goes to infinity and then we get the limit of the n goes to infinity.
00:37
Now am plus 1 because of the absolute value so we don't need to care about the minus 1 power n.
00:43
Therefore we have the x minus 3 power of n plus 1, dividing by the 2n plus 3.
00:50
That's an am plus 1.
00:52
Now we divide by an so we need to multiply by the reciprocal.
00:56
Therefore going to 2m plus 1, dividing by x minus 3 power of the n.
01:01
And now we see we can cancel the x minus 3 power and with this power here.
01:06
And then we can bring this one outside the limit.
01:10
Therefore we get absolute of the x minus 3 times the limit of the 2 n plus 1 divided by 2m plus 3 and goes to infinity.
01:22
And now because the degree on the top and the bottom there equal here, therefore this limit here equal to the 2 divided by 2 and 2...