00:01
Let a given let sn and tn be two sequence of, be two sequence, sequence in r such that, limit of, limit of sn is equals to s and limit of tn is equals to t in usual metric, in usual metric.
00:39
Then limit of, then limit of sn is equals to s for all epsilon is greater than 0, there is a, there is a n1 such that, such that for every, for every n is greater than, greater than or equal to n1, mod of sn minus s is less than epsilon by 2 is taken as equation 1 and limit of tn is equals to t for all epsilon is greater than 0, there is, there is n2 such that, such that for, for every, for every n is greater than or equal to n2, mod of, then mod of tn minus t is less than epsilon by 2 is taken as equation 2.
01:47
So, now let n3 is equals to maximum of, maximum of n1, n2 from 1 and 2, from 1, 2 we get, we get mod of sn minus s is less than epsilon by 2 and mod of tn minus t is less than epsilon by 2 for every, for every n is greater than or equal to n3 is taken as equation 3.
02:30
Now, l1 metric now, now l1 metric, l1 metric d of sn, sn comma tn comma s comma t is equals to, is equals to mod of sn minus s plus mod of tn minus t...