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I am trying to show that $\lim_{n\rightarrow \infty}\int_1^3\dfrac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and what its value is. I know that to do this I must show that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$ and that each $\dfrac{nx^{99}+5}{x^3+nx^{66}}$ is integrable on $[1,3]$ and the rest will follow. I am having a difficult time showing that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow x^{33}$ uniformly on $[1,3]$. So far my proof for uniform convergence is as follows.

Let $\epsilon > 0$, choose $N\in \mathbb{N}$ such that $\left|\frac{4}{N+1}\right|<\epsilon$. Then $n\geq N$ implies \begin{align*} \left|\frac{nx^{99}+5}{x^3+nx^{66}}-x^{33}\right|&=\left|\frac{nx^{99}+5-x^{33}(x^3+nx^{66})}{x^3+nx^{66}}\right|\\ &=\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\\ &\leq\left|\frac{4}{n+1}\right|\\ &\leq\left|\frac{4}{N+1}\right|<\epsilon \end{align*} and so we have uniform convergence on $[1,3]$.

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If we know $a_n(x)→a(x)$ on a compact interval like $I=[1,3]$, Doesn't this lead us to have uniformly convergent on $I$? –  B. S. Feb 28 '13 at 8:07
    
It is possible to attack the problem without using the term "uniform convergence." The $5$ part is quite harmless. And after you get rid of it, the $x^3$ at the bottom is not much of a problem when $n$ is large. –  André Nicolas Feb 28 '13 at 8:07
    
@AndréNicolas: You mean, we evaluate the integral and then take that limit? –  B. S. Feb 28 '13 at 8:08
    
I have updated the question with an attempt at completing the proof of uniform convergence. Is it correct? –  kaiserphellos Feb 28 '13 at 8:09
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Since $(nx^{99}+5)/(x^3+nx^{66})$ is uniformly bounded on $[1,3]$, we can apply Lebesgue's dominated convergence theorem to interchange the $\lim$ and $\int$. –  Frank Science Feb 28 '13 at 12:32
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2 Answers

You are on the good way, and to complete the proof you can by triangle inequality do: $$\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\leq \frac{5+x^{36}}{x^3+nx^{66}}\leq\frac{5+3^{36}}{1^3+n1^{66}}= \frac{5+3^{36}}{1+n}\rightarrow0,\quad\forall x\in[1,3]. $$

Hence, $\forall \epsilon>0,$ we can find $N\in\mathbb{N}$ such that $\forall x\in [1,3],\quad\forall n\geq N$ we have $\left|\frac{5-x^{36}}{x^3+nx^{66}}\right|\leq \epsilon. $ Now, you can conclude the uniform convergence.

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The integrand simplifies to $\dfrac{nx^{96}}{1+nx^{63}}+\dfrac{5}{x^3+nx^{66}}$.

The second term is less than $\dfrac{5}{n}$ on our interval, so it is harmless.

For the first term, divide. We get $x^{33}-\dfrac{x^{33}}{1+nx^{63}}$. The function $\dfrac{x^{33}}{1+nx^{63}}$ is less than $\dfrac{1}{n}$ on our interval.

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