Is the Function Uniformly Convergent on (0,1]?

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SUMMARY

The function defined as f_n(x)=\frac{nx}{nx+1} converges pointwise to 0 at x=0 and to 1 for all other x in [0,∞). However, it does not converge uniformly on the interval [0,1] as demonstrated by the limit calculation for x=1/n, which yields a supremum limit of 0.5, indicating that the supremum of |f_n(x)-1| does not approach 0. The discussion also addresses the uniform convergence on the interval (0,1], where the user seeks to prove that Limit_{n\rightarrow\infty} sup |f_n(x)-1|=0.

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

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

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

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

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

thanks for any help
 
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Your latex is screwed up.
 
Yeah do you know why that is?

thanks for any help
 
hmmmmm said:
Yeah do you know why that is?

thanks for any help

don't put TEX in capitals.
 

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