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Define a sequence $(f_n)$ of functions on $\Bbb{R}$ by putting $f_n(x) := \frac{x}{1+nx^2}$ for each natural number $n$ and for each real $x$. Let $f(x)$ be the limit of the sequence $(f_n(x))$ and $g(x)$ be the limit of the sequence $(f_n' (x))$ (the derivative of $(f_n(x))$). Find $f$ and $g$. Find all intervals $I$ where the sequences are uniformly convergent. For what $x$ is $f\,'(x) = g(x)$?

Here are my thoughts on the problem, from beginning to the end:

It seems that $\displaystyle\lim_{n\to\infty}\frac{x}{1+nx^2} = 0$ for all $x$, so $f(x) = 0$.

$$f_ n'(x) =\frac{1-nx^2}{(nx^2 +1)^2}\;,$$ so $$\lim_{n\to\infty}f_n'(x) = \lim_{n\to\infty}\frac{\frac1n-x^2}{nx^2+2x^2+\frac1n}= 0$$ as well.

So far it seems that $g(x) = f(x) = 0$.

When I look at solutions of these sort of problems it seems like the maxima of the sequences are found. By inspections of $(f_n'(x))$ we see that $(f_n'(x)) = 0$ when $1-nx^2 = 0$, i.e. when $x = \pm \frac1{\sqrt n}$. From plotting or a few calculations we see that the sequence takes its maxima in the positive solution.

Now, the standard procedure is apparently to insert the maxima into the absolute value $$\left|f\left(\frac1{\sqrt n}\right) - f_n\left(\frac1{\sqrt n}\right)\right| = \left|f_n\left(\frac1{\sqrt n}\right)\right| = \frac1{2\sqrt n} \to 0$$ when $n$ approaches infinity. From this we see that $f_n$ is uniformly convergent, although my textbook doesn't have any theorems or definitions that incorporate any suprema or maxima (I'd greatly appreciate some insight in this matter).

Similar calculations give that $(f_n'(x))$ takes its maxima in $\sqrt{\frac3{n}}$, but here $|f_n'(x)| = \frac18$, so $f_n'(x)$ is not uniformly convergent.

The last question is the one that raised a warning flag. Assuming all my calculations are correct, $f'(x) = g(x)$ for all $x$ since both are the zero function. I think it is suspicious how trivial my answer is.

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Please check to be sure that I correctly converted your ASCII to $\LaTeX$. – Brian M. Scott Mar 9 '12 at 11:42
This is correct, thank you. – user26334 Mar 9 '12 at 11:47
Sorry about that extra parenthesis: you caught it just before I did. – Brian M. Scott Mar 9 '12 at 11:49
You may want to check the special case $x=0$ – user20266 Mar 9 '12 at 12:51
Right, so the limit of $(f_n' (0))$ is 1 The limit if $(f_n (0))$ is still $0$. The answer to the last question now becomes that they are equal everywhere except for the origin. – user26334 Mar 9 '12 at 13:05

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