Jtdylktuy

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

Dividing by $n x^q$ so we have
$$dfrac{n x^p+x^q}{x^p+n x^q}=dfrac{x^{p-q}+1/n}{(1/n) x^{p-q}+1}leq dfrac{x^{-1}+1/n}{(1/n) x^{p-q}+1}leq x^{-1}+1/n quadtext{ since } 0leq xleq 1 $$
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.

Writing $f_ncolon xin(0,1]mapsto frac{n x^p+x^q}{n x^q+x^p}$

• pointwise convergence to $fcolon xin(0,1]mapsto x^{p-q}$ (with $p-q > -1$), which is integrable on $(0,1]$.


• for all $ngeq 1$ and $xin(0,1]$,
  $$
  0 leq f_n(x) leq frac{n x^p+x^q}{n x^q} = frac{1}{x^{q-p}}+frac{1}{n} leq frac{1}{x^{q-p}}+1= g(x)
  $$
  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 $int_{(0,1]} f_n xrightarrow[ntoinfty]{}int_{(0,1]} f = frac{1}{p-q+1}$.

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:
$$int_{0}^{1}frac{nx^p+x^q}{nx^q+x^p},dx sim int_{0}^{1}x^{p-q},dx = frac{1}{p-q+1}.$$