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I am given [tex]f_n(x)=\frac{nx}{nx+1}[/tex] defined on [tex] [0,\infty) [/tex] and I have that the function converges pointwise to [tex] 0 \ \mbox{if x=0 and} 1\ \mbox{otherwise}[/tex]

Is the function uniform convergent on [tex] [0,1] [/tex]?

No. If we take x=1/n then [tex]Limit_{n\rightarrow\infty}|\frac{1/n*n}{1+1/n*n}-1|=0.5[/tex]

which implies that [tex]Limit_{n\rightarrow\infty} sup |f_n(x)-1|[/tex] is not 0.

I am then asked if it converges uniformly on the interval [tex](0,1][/tex] which I think it does but how do I show that [tex]Limit_{n\rightarrow\infty} sup |f_n(x)-1|[/tex]=0?

thanks for any help

Is the function uniform convergent on [tex] [0,1] [/tex]?

No. If we take x=1/n then [tex]Limit_{n\rightarrow\infty}|\frac{1/n*n}{1+1/n*n}-1|=0.5[/tex]

which implies that [tex]Limit_{n\rightarrow\infty} sup |f_n(x)-1|[/tex] is not 0.

I am then asked if it converges uniformly on the interval [tex](0,1][/tex] which I think it does but how do I show that [tex]Limit_{n\rightarrow\infty} sup |f_n(x)-1|[/tex]=0?

thanks for any help

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