Finding the limit of this integral: $\lim_{n\to\infty} \int_0^1 \frac{n x^p+x^q}{x^p+n x^q} dx$ if $q<p+1$

Question

I am trying to find the following limit provided: $q<p+1$: $$ \lim_{n\to\infty} \int_0^1 \dfrac{n x^p+x^q}{x^p+n x^q} dx$$

Dividing by $n x^q$ so we have $$\dfrac{n x^p+x^q}{x^p+n x^q}=\dfrac{x^{p-q}+1/n}{(1/n) x^{p-q}+1}\leq \dfrac{x^{-1}+1/n}{(1/n) x^{p-q}+1}\leq x^{-1}+1/n \quad\text{ since } 0\leq x\leq 1 $$ or maybe $$\dfrac{n x^p+x^q}{x^p+n x^q}\leq \dfrac{n x^{q-1}+x^q}{x^p+n x^{p+1}} $$

I am trying to use MCT or DCT in somehow, or maybe other things.

Please help me solving this problem I am preparing for a prelim exam in January.

Answer 1

Writing $f_n\colon x\in(0,1]\mapsto \frac{n x^p+x^q}{n x^q+x^p}$

• pointwise convergence to $f\colon x\in(0,1]\mapsto x^{p-q}$ (with $p-q > -1$), which is integrable on $(0,1]$.
• for all $n\geq 1$ and $x\in(0,1]$, $$ 0 \leq f_n(x) \leq \frac{n x^p+x^q}{n x^q} = \frac{1}{x^{q-p}}+\frac{1}{n} \leq \frac{1}{x^{q-p}}+1= g(x) $$ where $g$ is integrable on $(0,1]$ as $q-p < 1$.

Then, unless I have forgotten something you can apply the DCT to get that $\int_{(0,1]} f_n \xrightarrow[n\to\infty]{}\int_{(0,1]} f = \frac{1}{p-q+1}$.

Answer 2

Hint: $$\frac{nx^p+x^q}{nx^q+x^p}-x^{p-q}=\frac{x^q-x^{2p-q}}{nx^q+x^p}.$$ If you prove that the last term is small, then: $$\int_{0}^{1}\frac{nx^p+x^q}{nx^q+x^p}\,dx \sim \int_{0}^{1}x^{p-q}\,dx = \frac{1}{p-q+1}.$$