Let fn = \frac{1}{n(1+x^2)}. Find the pointwise limit of fn. Is the convergence uniform?

If I'm not mistaken, fn \rightarrow 0, For all x in R.

I believe that for this function the convergence is uniform. By letting N> \frac{1}{\epsilon} we have,

\vert 0  - \frac{1}{n(1+x^2)} \vert < \vert \frac {1}{n} \vert < \frac{1}{N} < \epsilon.

So our choice of N, only depends on epsilon, for all x in R.




Let fn = \frac{x^2+nx}{n}. Find the pointwise limit of fn. Is the convergence uniform?

Now fn \rightarrow f(x) = x , for all x in R.

In this case however, there is no uniform convergence because:

\vert \frac{x^2+nx}{n} -x \vert = \frac{x^2}{n}

So our choice of N would imply taking N > \frac{x^2}{\epsilon}. So our choice of N depends on both epsilon and x. Could it be that for a closed interval of the form [a,b] the convergence is uniform?



Thank you