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Consider the functions $f_n:[-1,1]\to\mathbb{R}$ defined by $$f_n(x):= \frac{x}{\sqrt{x^2 + \tfrac 1n}}$$ and determine whether the convergence is uniform or pointwise.

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We have that

If a sequence of functions $(f_n)_{n\geq 1}$ converges pointwise to a function $f$ and uniformly to a function $g$, then $f=g$.

and

If a sequence of continuous functions $(f_n)_{n\geq 1}$ converges uniformly to a function $f$, then $f$ is continuous.

  • Prove convergence pointwise to the function $f\colon[-1,1]\mapsto \operatorname{sign}(x)$ as you started: fix $x_0\in[-1,1]$ arbitrarily, and consider the limit of the sequence of real numbers $(f_n(x_0))_{n\geq 1}$. (You may want to consider separately the case $x_0<0$, $x_0=0$ and $x_0>0$.)

  • Then, for uniform convergence... see how to apply the two statements above.

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  • $\begingroup$ A trick to avoid one distinction of case (but that's really not necessary, and the "optimization of time" is marginal): all the $f_n$'s is odd, so you can restrict yourself to the study on $[0,1]$. $\endgroup$ – Clement C. Feb 13 '16 at 14:06
  • $\begingroup$ Fix for instance $x_0 \in(0,1]$. Then $$ \frac{x}{\sqrt{x^2+\frac{1}{n}}} = \frac{x}{\lvert x\rvert}\cdot\frac{1}{\sqrt{1+\frac{1}{nx_0}}} = 1\cdot \frac{1}{\sqrt{1+\frac{1}{nx_0}}}. $$ Since $nx_0 \xrightarrow[n\to\infty]{} \infty$, the RHS converges to $1$. You do not need $\varepsilon-\delta$-proofs here: if you have a continuous function $\phi$ and a sequence $a_n\to \ell$, then $\phi(a_n) \to \phi(\ell)$. $\endgroup$ – Clement C. Feb 13 '16 at 14:10
  • $\begingroup$ Sorry, in the comment above the $x$'s should be $x_0$'s. (Typing $\LaTeX$ in comments is a bit painful.) $\endgroup$ – Clement C. Feb 13 '16 at 14:16
  • $\begingroup$ Fix $\varepsilon> 0$. We can write $$ \left\lvert \frac{1}{\sqrt{1+\frac{1}{nx_0}}} - 1\right\rvert = \left\lvert \frac{1-\sqrt{1+\frac{1}{nx_0}}}{\sqrt{1+\frac{1}{nx_0}}}\right\rvert = \frac{\sqrt{1+\frac{1}{nx_0}}-1}{\sqrt{1+\frac{1}{nx_0}}} \leq \sqrt{1+\frac{1}{nx_0}}-1 $$ Having the RHS less than $\varepsilon$ is equivalent to having $$ \sqrt{1+\frac{1}{nx_0}} \leq 1+\varepsilon $$ which, reorganizing and massaging a bit, is equivalent to having $$ \frac{1}{n} \leq x_0\left( (1+\varepsilon)^2-1 \right). $$ $\endgroup$ – Clement C. Feb 13 '16 at 14:32
  • $\begingroup$ So picking $N = \left\lceil \frac{1}{x_0\left( (1+\varepsilon)^2-1 \right)} \right\rceil$ ensures that, for $n\geq N$, the RHS is less than $\varepsilon.$ (Having checked carefully the derivation, but that's the gist of it.) $\endgroup$ – Clement C. Feb 13 '16 at 14:32

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