Question

1. using the definition show that lim n/2n-3 =1/2 (the limit is as n approches to infinity).

Answer 1

definition means that given any \(\epsilon >0\) there is an \(N_\epsilon\) such that if \(n>N\) you have
\[|\frac{n}{2n-3}-\frac{1}{2}|<\epsilon\]

Answer 2

you work backwards to find the \(N\)

first algebra inside the absolute value signs, then it will be clear what \(N\) you need to pick, depending on \(\epsilon\)

Answer 3

u r right, it is using epsilon..

Answer 4

lol abusive because i did not show the necessary steps?
diy

Answer 5

1. using the definition show that lim n/2n-3 =1/2 (the limit is as n approches to infinity).

Answer 6

\[\lim_{n \rightarrow \infty} \frac{ n }{ 2n-3 }\]
Dividing Nr. and Dr. by n we find
\[= \lim_{n \rightarrow \infty} \frac{ 1 }{ 2-3(\frac{ 1 }{ n })}\]
\[n \rightarrow \infty, \frac{ 1 }{ n }\rightarrow 0\]
Therefore
\[= \frac{ 1 }{ 2-0}\]
\[= \frac{ 1 }{ 2}\]
Thus proved

Answer 7

definition means that given any ϵ>0 there is an
N
ϵ such that if
n>N you have

Answer 8

using the definition show that lim n/2n-3 =1/2 (the limit is as n approaches to infinity).definition means that given any ϵ>0 there is an N ϵ such that if n>N

Answer 9

show that of summation of 1/n2^n from 1 to infinity = ln2