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
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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