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

If I'm not mistaken, f_{n}$\displaystyle \rightarrow$ 0, For all x in R.

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

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

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

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

Now f_{n}$\displaystyle \rightarrow f(x) = x $, for all x in R.

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

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

So our choice of N would imply taking N > $\displaystyle \frac{x^2}{\epsilon}$. 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