Proof of pointwise and uniform convergence I was wondering if someone could check my proof of pointwise and uniform convergence for the sequence $f_n=\bigl((1/n)+x^2\bigr)^{1/2}$. By taking the limit of $f_n$ as $n$ goes to infinity, we see that the pointwise limit is the function $f(x)=(x^2)^{1/2}=|x|$.
Claim:The sequence $\{f_n\}$ converges uniformly to $f$.
Proof: Let $\epsilon>0$. Choose $N$ such that $1/(N)^{1/2}<\epsilon$. Let $n\geq N$ and $x$ be real. Then,
$$|f_n(x)-f(x)|=\Bigl|((1/n)+x^2)^{1/2}-(x^2)^{1/2}\Bigr|$$
Multiplying by its conjugate yields
$$\Bigl|1/\bigl[n\bigl((1/n)+x^2\bigr)^{1/2}+(x^2)^{1/2})\bigr)\bigr]\Bigr|\leq 1/(n)^{1/2}<=1/(N)^{1/2}<\epsilon.$$
 A: The rhs in your first displayed equation, times it's conjugate, is simply
$$\left|\left( (\frac{1}{n})+x^2 \right)^{\frac{1}{2}}-\left(x^2\right)^{\frac{1}{2}}\right| \times \left|\left( (\frac{1}{n})+x^2 \right)^{\frac{1}{2}}+\left(x^2\right)^{\frac{1}{2}}\right| = \left( (\frac{1}{n})+x^2 \right)-x^2=\frac{1}{n}$$
This will in fact allow you to prove uniform convergence (for large $x$, see below).
The term you've got in the lhs of your last displayed equation 
$$|1/[n((1/n)+x^2)^{1/2}+ (x^{2})^{1/2})]|$$
seems to be very strange to me, even the $x$-term appears somehow in the denominator. Maybe it is due to a typo?
If you multiply with the conjugate 
$$\left|\left( (\frac{1}{n})+x^2 \right)^{\frac{1}{2}}+\left(x^2\right)^{\frac{1}{2}}\right| $$
(as you call it), as mentioned earlier in my comment, you have to be aware of the fact that this is $<1$ for small $x$ and large $n$. In that case, when multiplying the rhs of your inequality, the inequality is not preserved. This means, for small $x$ you need a different reasoning.
