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$f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ converges pointwise in a set $E = [0, \infty)$ to $f(x) = x$.

This problem reminds me a lot of how $\frac{x}{n}$ fails to converge uniformly to $f=0$ on $[0,\infty]$, but does converge uniformly on $[0,a]$. However, when we proved that we used the definition of uniform convergence. I do not wish to do that here.

I wish to use the Weierstrass test here. And that is where I'm stuck. I cannot find $M_n = \sup\limits_{x \in E} |f_n(x)-f(x)|$. Can someone help me out here?

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    $\begingroup$ Hint: $\sqrt{x^2 + c^2} \le x+c$ for $x, c \ge 0$. $\endgroup$ – achille hui May 9 '15 at 8:15
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Observe that $$ 0<f_n(x)-x=\sqrt{x^2+\frac{1}{n^2}}-x=\frac{x^2+\frac{1}{n^2}-x^2}{\sqrt{x^2+\frac{1}{n^2}}+x}=\frac{1}{n^2\sqrt{x^2+\frac{1}{n^2}}+n^2x}\le\frac{1}{n}, $$ as $$ n^2\sqrt{x^2+\frac{1}{n^2}}+n^2x\ge n^2\cdot \frac{1}{n}=n. $$

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Hint: $$\sqrt{x^2}<\sqrt{x^2+\frac1{n^2}}\le\sqrt{x^2+2x\frac1n+\frac1{n^2}} $$

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