<s>[INST] Prove that $\lim\limits_{n\rightarrow \infty}\int_1^3\frac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and evaluate it. 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 $[1,3]$.
 [/INST] 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. 
</s>