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Calculate $\text{lim}_{n\rightarrow\infty} \displaystyle\int_{0}^{1} \displaystyle\frac{x}{1+n^2x^2}$

Using the Dominant Convergence Theorem which has three steps to show to achieve: $\int f=\int\text{lim} f_n = \text{lim}\int f_n$

  1. Show $f_n \in L^1$ i.e. show it is Riemann Integrable

  2. $|f_n|\leq g \in L^1$

  3. $f_n \rightarrow f$ almost everywhere

So far I have:

$f_n(x) = \displaystyle\frac{x}{1+n^2x^2}$ on the interval $[0,n]$.

  1. $f_n \in L^1$ since it is continuous on $[0,n]$ therefore it is Riemann Integrable

$\int f_n = \int_{0}^{n} \displaystyle\frac{x}{1+n^2x^2} dx = \frac{1}{2n^2} log(1+n^4)$

I am stuck on bounding $|f_n|$.

Any help is appreciated.

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  • $\begingroup$ I wonder why it is asked to apply the DCT, while $\frac{x}{1+n^2 x^2}\in\left[0,\frac{1}{2n}\right]$ for any $x\in[0,1]$ already ensures that the limit is zero. $\endgroup$ – Jack D'Aurizio Mar 30 '18 at 15:01
  • $\begingroup$ it was the direction in the textbook so that is what i was trying to use @JackD'Aurizio $\endgroup$ – rover2 Apr 2 '18 at 22:27
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Computing the derivative of $f_{n}$ at the point where it vanishes is $1/n$, and so $\max_{x\in[0,1]}f_{n}(x)\leq\max\{f_{n}(0),f_{n}(1),f_{n}(1/n)\}=\max\{0,1/(2n),1/(1+n^{2})\}=1/(2n)\leq 1$, just take $g=1$.

Note that here we consider $L^{1}[0,1]$, so $g\in L^{1}[0,1]$.

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Why are you writing $[0,n]$ instead of $[0,1]$? Just use the bound $\frac x {1+n^{2}x^{2}} \leq \frac 1 {2n} \leq 1$. (This just rewriting of the inequality $(nx-1)^{2} \geq 0$).

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