let us start with, lim m → ∞ x m = ( x 1 , x 2 , x 3 , . . . . . . . . . , x m ) \lim_{m \to \infty} {x_m} = (x_1,x_2,x_3, . . . . . . . . . , x_m) lim m → ∞ x m = ( x 1 , x 2 , x 3 , ......... , x m ) = {x}
for ϵ > 0 , \epsilon > 0, ϵ > 0 , n 0 ∈ N n_0 \isin \N n 0 ∈ N such that
∣ ∣ x n − x ∣ ∣ < ϵ || {x_n} - {x} || < \epsilon ∣∣ x n − x ∣∣ < ϵ for ∀ \forall ∀ n ≥ n 0 n \geq n_0 n ≥ n 0
so n ≥ n 0 n \geq n_0
n ≥ n 0 , we can say that
∣ ∣ ( x 1 m , x 2 m , x 3 m , . . . . . . . . . , x n m ) − ( x 1 , x 2 , x 3 , . . . . . . , x n ) ∣ ∣ < ϵ ||(x_{1m},x_{2m}, x_{3m}, . . . . . . . . . ,x_{nm}) - ( x_1,x_2,x_3,......,x_n )|| < \epsilon ∣∣ ( x 1 m , x 2 m , x 3 m , ......... , x nm ) − ( x 1 , x 2 , x 3 , ...... , x n ) ∣∣ < ϵ
∣ ∣ ( x 1 m − x 1 ) , ( x 2 m − x 2 ) , . . . . . ( x n m − x n ) ∣ ∣ < ϵ || (x_{1m}-x_1),(x_{2m}-x_2),.....(x_{nm}-x_n)|| < \epsilon ∣∣ ( x 1 m − x 1 ) , ( x 2 m − x 2 ) , ..... ( x nm − x n ) ∣∣ < ϵ
( x 1 m − x 1 ) 2 + ( x 2 m − x 2 ) 2 + . . . . . ( x n m − x n ) 2 < ϵ \sqrt { (x_{1m}-x_1)^2+(x_{2m}-x_2)^2+.....(x_{nm}-x_n)^2} < \epsilon ( x 1 m − x 1 ) 2 + ( x 2 m − x 2 ) 2 + ..... ( x nm − x n ) 2 < ϵ
( x i m − x i ) 2 < ϵ \sqrt{{(x_{im} - x_i)}^2} < \epsilon ( x im − x i ) 2 < ϵ
∣ ∣ ( x i m − x i ) ∣ ∣ < ϵ ||{(x_{im} - x_i)} ||< \epsilon ∣∣ ( x im − x i ) ∣∣ < ϵ
lim x → ∞ x i m = x i \lim_{x \to \infty} {x_{im}} = x_i lim x → ∞ x im = x i
Conversely, assume that lim x → ∞ x i m = x i \lim_{x \to \infty} {x_{im}} = x_i lim x → ∞ x im = x i
since, for ϵ > 0 , \epsilon > 0, ϵ > 0 , ∃ \exists ∃ n 0 i ∈ N n_{0i} \isin \N n 0 i ∈ N such that
∣ ∣ x i m − x i ∣ ∣ < ϵ || x_{im} - x_i|| < \epsilon ∣∣ x im − x i ∣∣ < ϵ for every value of n > n 0 i n > n_{0i} n > n 0 i (1)
for each i = 0 , 1 , 2 , 3 , 4 , . . . . . . . . , n i=0,1,2,3,4,. .......,n i = 0 , 1 , 2 , 3 , 4 , ........ , n
Consider the maximum of ( x 01 , x 02 , . . . . . . . . , x 0 n ) (x_{01},x_{02},........,x_{0n}) ( x 01 , x 02 , ........ , x 0 n )
then (1) will hold true simultaneously for each i.
∣ ∣ x i m − x i ∣ ∣ < ϵ || x_{im} - x_i || < \epsilon ∣∣ x im − x i ∣∣ < ϵ ∀ \forall ∀ n ≥ n 0 i ∣ m a x n\geq n_{0i}|_{max} n ≥ n 0 i ∣ ma x
( x i m − x i ) 2 < ϵ 2 (x_{im}-x_i)^2 < \epsilon^2 ( x im − x i ) 2 < ϵ 2 for each i, n ≥ n 0 i ∣ m a x n\geq n_{0i}|_{max} n ≥ n 0 i ∣ ma x
( x 1 m − x 1 ) 2 + ( x 2 m − x 2 ) 2 + . . . . . . + ( x n m − x n ) 2 < n ϵ 2 (x_{1m}-x_1)^2+(x_{2m}-x_2)^2+......+(x_{nm}-x_n)^2 < n\epsilon^2 ( x 1 m − x 1 ) 2 + ( x 2 m − x 2 ) 2 + ...... + ( x nm − x n ) 2 < n ϵ 2
∣ ∣ x m − x ∣ ∣ < n ϵ || x_m - x || < \sqrt{n}\epsilon ∣∣ x m − x ∣∣ < n ϵ for each values of i, n ≥ n 0 i ∣ m a x n\geq n_{0i}|_{max} n ≥ n 0 i ∣ ma x
lim m → ∞ x m = \lim_{m \to \infty} {x_m} = lim m → ∞ x m = {x}
Hence proved
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