Limits and Measure Theory How would I evaluate these limits?
$$ \lim_{n \to \infty} \int_0^\infty \frac{n}{1+(nx)^2} \ dx$$
and
$$ \lim_{n \to \infty} \int_0^\infty \frac{(1+(nx)^2)}{(1+nx^2)^n} \ dx$$
 A: As already noted, the integrals considered in the first case do not depend on $n$ hence their limit is not very mysterious.
Let $I_n$ denote the integral considered in the second case. By the change of variable $u=nx$, 
$$
nI_n=\displaystyle\int_0^\infty(1+u^2)f_n(u)\mathrm{d}u,\qquad\mbox{with}\ f_n(u)=(1+u^2/n)^{-n}.
$$
The sequence of positive functions $(f_n)_n$ is nonincreasing (*) and $f_n\to f$ pointwise when $n\to\infty$, with $f(u)=\mathrm{e}^{-u^2}$. Since $f_2\ge f_n$ for very $n\ge2$ and $I_2$ is finite, by Lebesgue dominated convergence theorem, 
$$
nI_n\to J=\displaystyle\int_0^\infty(1+u^2)f(u)\mathrm{d}u=1+(3\pi/2).
$$ 
In particular, $I_n\to0$ when $n\to\infty$.
(*) This is implied by the fact that, for every $c\ge0$, the function $z\mapsto(1+c/z)^z$ is nondecreasing on $z\ge0$. To show this last fact, differentiate twice the logarithm of this function.
Added later on Following Shai's suggestion, I mention that the proof above shows that the sequence $(nI_n)_n$ is nonincreasing. For instance $nI_n\le2I_2$ for every $n\ge2$ (note that $I_1=+\infty$ but that $I_2$ is finite).
A: 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 $0$.
A: Actually, the second integral, like the first, could have been a "calc 1 problem" (but it is useful to practice the dominated convergence theorem). 
Indeed, the inequality $\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }} \le 1$ (also used by bobobinks) gives
$$
\int_0^\varepsilon  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x}  \le \varepsilon, 
$$
for any $\varepsilon > 0$ (arbitrarily small, but fixed). On the other hand, for any $n \geq 4$, 
$$
\int_\varepsilon ^\infty  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x}  \le \int_\varepsilon ^\infty  {\frac{{\varepsilon ^{ - 2} x^2  + n^2 x^2 }}{{(2\sqrt n x)^n }}\,{\rm d}x}  \le \frac{{\varepsilon ^{ - 2}  + n^2 }}{{(2\sqrt n )^n }}\int_\varepsilon ^\infty  {x^{2 - n} \,{\rm d}x} = \frac{{\varepsilon ^{ - 2}  + n^2 }}{{(2\sqrt n )^n }}\frac{{\varepsilon ^{3 - n} }}{{n - 3}}. 
$$
Since, for $\varepsilon > 0$ fixed, the expression on the right-hand side tends to $0$ as $n \to \infty$, it follows that $\int_0^\infty  {\frac{{1 + n^2 x^2 }}{{(1 + nx^2 )^n }}\,{\rm d}x} \to 0$ too.
A: Limit for Intergral from 0 to infinity of [1-(cosx to power n)] over 1+(x to the power 2) of dx
$$
\lim_{n\to\infty}\int_0^\infty\frac{1-\cos^n(x)}{1+\cos^n(x)}\,\mathrm{d}x
$$
