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Test whether or not $(f_n(x)) = \frac{nx}{n+x}$ on $I=[0,1]$ converges uniformly on $I=[0,1]$.

My attempt:

$\displaystyle \lim_{n \to \infty}f_n(x) = \lim_{n \to \infty}\frac{nx}{n+x} = \lim_{n \to \infty} \frac{x}{1 + \frac{x}{n}} = x$ for all $x \in I$.

Thus $(f_n)$ converges pointwise to $f(x) = x$ on $I$.

For uniform convergence, we must test the pointwise limit for uniform convergence:

$$||f_n - x|| = \sup\bigg\{ \bigg|\frac{nx}{n + x} -x\bigg| : x\in I \bigg\} = \sup \bigg\{\bigg| \frac{-x^2}{n+x}\bigg|: x \in [0,1] \bigg\} = \frac{1}{n+1}$$

Notice however, that $\frac{1}{n+1} \to 0$ as $n \to \infty$ and hence $(f_n)$ converges uniformly on $I$.

Is this correct?

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  • $\begingroup$ It looks flawless to me. $\endgroup$ – Jack D'Aurizio May 14 '15 at 15:46
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Thats very correct. Given any $\epsilon >0$ take $N(\epsilon) $ as $[\frac{1}{\epsilon}]+1$ and the definition follows.

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