In a hyper-real context, uniform convergence allows for a simpler description. So, if is infinitely close to for all x in the space of and all infinite m, a sequence converges to f uniformly.(For a related concept of uniform continuity, see micro continuity)
f ( x ) = n x 2 + 1 n x + 1 f(x)=\frac{nx^2+1}{nx+1} f ( x ) = n x + 1 n x 2 + 1
lim x → ∞ f n ( x ) = lim x → ∞ [ n x 2 + 1 n x + 1 ] = lim x → ∞ [ n n x 2 + 1 n n x n + 1 n ] = lim x → ∞ [ x 2 + 1 n x + 1 n ] = lim x → ∞ [ x 2 + 1 ∞ x + 1 ∞ ] = x \lim\limits_{x \to \infin}f_n(x)=\lim\limits_{x \to \infin}[\frac{nx^2+1}{nx+1}]=\lim\limits_{x \to \infin}[\frac{\frac{n}{n}x^2+\frac{1}{n}}{\frac{nx}{n}+\frac{1}{n}}]=\lim\limits_{x \to \infin}[\frac{x^2+\frac{1}{n}}{x+\frac{1}{n}}]=\lim\limits_{x \to \infin}[\frac{x^2+\frac{1}{\infin}}{x+\frac{1}{\infin}}]= x x → ∞ lim f n ( x ) = x → ∞ lim [ n x + 1 n x 2 + 1 ] = x → ∞ lim [ n n x + n 1 n n x 2 + n 1 ] = x → ∞ lim [ x + n 1 x 2 + n 1 ] = x → ∞ lim [ x + ∞ 1 x 2 + ∞ 1 ] = x
When x=0
f n ( x ) = 0 f_n(x)=0 f n ( x ) = 0
Therefore, f ( x ) = [ x i f x ≠ 0 0 i f x = 0 ] f(x)={x \space if \space x \not =0\brack 0 \space if \space x=0} f ( x ) = [ 0 i f x = 0 x i f x = 0 ]
lim x → ∞ f n ( x ) = f ( x ) \lim\limits_{x \to \infin}f_n(x)= f(x) x → ∞ lim f n ( x ) = f ( x )
To prove that the above convergence is not pointwise
Suppose that f n → f f_n \to f f n → f
Therefore, ∃ ε > 0 , ∃ x > [ 1 , 2 ] , ∀ N ϵ N , ∃ n > N \exist \varepsilon >0, \exist x>[1,2], \forall N \epsilon \N, \exist n> N ∃ ε > 0 , ∃ x > [ 1 , 2 ] , ∀ N ϵ N , ∃ n > N
∣ f n ( x ) − f ( x ) ∣ < ε |f_n(x)-f(x)|< \varepsilon ∣ f n ( x ) − f ( x ) ∣ < ε
Now , taking ε = 1 10 , x = 1 n \varepsilon = \frac{1}{10}, x=\frac{1}{\sqrt{n}} ε = 10 1 , x = n 1
Therefore, ∣ n x 2 + 1 n x + 1 − x ∣ = 1 − x n x + 1 > ε = 1 10 |\frac{nx^2+1}{nx+1}-x|=\frac{1-x}{nx+1}>\varepsilon=\frac{1}{10} ∣ n x + 1 n x 2 + 1 − x ∣ = n x + 1 1 − x > ε = 10 1
Which is a contradiction to the fact that f n ( x ) f_n(x) f n ( x )
Hence , f n ( x ) f_n(x) f n ( x )
