Pog i will give u a pog if u get it correct

Mathematics

2 answers:

Eva8 [605]3 years ago

4 0

Answer:

WHOS YOUR FAV MCYT

Step-by-step explanation: WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT WILBUR SOOT TOMMYINNIT

Rina8888 [55]3 years ago

3 0

Answer:

32.4

Step-by-step explanation:

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Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .