Find $\lim_{n \rightarrow \infty} \int_{0}^{1} \frac{n^3 x^{3/4}}{ 1 + n^4 x^2}.$ Find $$\lim_{n \rightarrow \infty} \int_{0}^{1} \frac{n^3 x^{3/4}}{ 1 + n^4 x^2}.$$ 
My attempt:
1-I am thinking about using monotone convergence theorem. but I do not understand what is the meaning of uniformly in the statement of the bounded convergence theorem given below? 



Could anyone explains this for me, please?
2- For my check for the conditions of The bounded convergence theorem:
First: I know that $\{ f_{n} = \frac{n^3 x^{3/4}}{ 1 + n^4 x^2}\}$ is measurable as it is continuous on its domain and as the domain is the set $[0,1]$ which is measurable and its measure is 1, hence this set has a finite measure.
Second: and I know also that $\{f_{n}\}$ is bounded by 1 on $[0,1]$ for all $n.$
Third: I know that the pointwise limit is $o$, then  $$\lim_{n \rightarrow \infty} \int_{0}^{1} \frac{n^3 x^{3/4}}{ 1 + n^4 x^2} = \int_{[0,1]} o. $$
And now $f$ is a simple function as it is measurable and of one finite value, which is $0$ and since $m{[0,1]}=1 $ then $\int_{[0,1]} f = o.1 =0. $ 
Is my checking for the conditions of the theorem correct? and is my final answer correct?  
 A: To apply Bounded Convergence Theorem you need the condition that for some $C$, $|f_n(x)| \leq C$ for all $n$ and for all $x$. 
This condition is not satisfied here: if $n^{3}x^{3/4} \leq C (1+n^{4}x^{2})$ for all $n$ and for all $x$ then we can put $x =\frac 1 {n^{2}}$ to get $n^{1.5} \leq 2C$ for all $n$ which is false.  Hence Bounded Convergence Theorem cannot be applied.
You can apply Dominated Convergence Theorem where the constant $C$ is replaced by a function of $x$ which is integrable. Note that $ \frac {n^{3}x^{3/4}} {1+n^{4}x^{2}}\leq \frac {n^{3}x^{3/4}} {n^{4}x^{2}} \leq x^{-5/8}$ which is integrable. Hence we can apply Dominated Convergence Theorem and conclude that the given limit exists and is $0$. 
Here is an elementary proof which does not use DCT:
Split the integral into integral from $0$ to $n^{-2/5}$ and the integral from $n^{-2/5}$ to $1$. In the first term use the bound $x^{-5/8}$ mentioned above and calculate the integral explcitly. In the second term verify that $\frac {n^{3}x^{3/4}} {(1+n^{4}x^{2})} \leq \frac {n^{3}x^{3/4}} {n^{4}x^{2}} \leq \frac  1 {\sqrt n}$. 
A: Let $n^2x=y \Rightarrow dx=\frac{dy}{n^2}$, then 
$$I =\lim_{n \rightarrow\infty} \int_{0}^{\infty} \frac{{y}^{3/4}}{\sqrt{n} (1+y^2)} dy=0.$$ as $$J=\int_{0}^{\infty} \frac{y^{3/4}}{1+y^2} dy ~is ~finte ~ and ~nonzero.$$  Because $$J \sim \int_{`}^{\infty} \frac{dy}{y^\beta} < \infty ~if ~ \beta >1$$. Here $\beta =5/4.$
Also use $y=\tan t$ , then by $\beta$-integral and Gamma functions we have  $$J=\int_{0}^{\infty} \frac{y^{3/4}}{1+y^2} dy= \int_{0}^{\pi/2} \tan ^{3/4} t dt =\frac{\Gamma(7/8) \Gamma(1/8)}{2}= \frac{\pi}{2} \csc(\pi/8).$$
