<s>[INST] Limits and Measure Theory How would I evaluate these limits?
$$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$
and
$$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$
 [/INST] From $(1+a)^n\geq1+na\;\;\;\forall a\geq0$ with $a=nx^2$, you have:
$\frac{1+n^2x^2}{(1+nx^2)^n}\leq1$. 
Furthermore, you have:$\frac{1+n^2x^2}{(1+nx^2)^n}<\frac{1+n^2x^2}{(nx^2)^n}=\frac{1}{n^nx^{2n}}+\frac{1}{n^{n-2}x^{2(n-1)}}<\frac{2}{x^2}$ if $n\geq2$ and $x>0$.
Hence, by the Lebesgue dominated convergence theorem, you need only consider the integral of the limit function to evaluate the limit 
(let $g(x)=\begin{cases}
1 & \text{ if } 0\leq x\leq1 \\ 
 \frac{2}{x^2}& \text{ if } x>1 
\end{cases}$ ).
Now $0\leq\frac{1+n^2x^2}{(1+nx^2)^n}\leq\frac{n^2+n^2x^2}{(1+x^2)^n}=\frac{n^2}{(1+x^2)^{n-1}}$.
Hence $lim_{n\rightarrow\infty}\frac{1+n^2x^2}{(1+nx^2)^n}=0$ by the squeeze theorem (we can neglect $x=0$ since it constitutes a set of measure zero).  
Hence the desired integral is equal to $0$.
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