Question #310728

Let fn(x)= nx/(1+nx) is not uniformly convergent on [0,1]

1
Expert's answer
2022-03-14T17:00:52-0400

Solution


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For x=0x=0 ,


limnfn(x)=limnnx1+nxlimnfn(0)=limnn(0)1+n(0)=0\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{1 + nx}}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( 0 \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( 0 \right)}}{{1 + n\left( 0 \right)}}\\ = 0


For 0 < x \le 1\


limnfn(x)=limnnx1+nxlimnfn(x)=limnnxn(1n+x)=limnx1n+x=1\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{1 + nx}}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{n\left( {\frac{1}{n} + x} \right)}}\\ = \mathop {\lim }\limits_{n \to \infty } \frac{x}{{\frac{1}{n} + x}}\\ = 1


Since we can see that the two limits are not equal, therefore fn(x)=nx1+nx{f_n}\left( x \right) = \frac{{nx}}{{1 + nx}} , is not uniformly convergent over the interval [0,1][0, 1] .

 

 


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