If $a>0$ and $b>1$ and $f:\left[0,1\right] \rightarrow \mathbb{R}$. Then value of $\displaystyle \lim_{n\rightarrow \infty}n^{\frac{a}{b}}\int^{1}_{0}\frac{f(x)}{1+n^{a}x^{b}}dx$ is
What i try
For $0\leq a<1$
$$I =\lim_{n\rightarrow \infty}n^{\frac{a}{b}}\int^{1}_{0}f(x)dx$$
For $a=1$, we have $$I=\lim_{n\rightarrow \infty}\int^{1}_{0}f(x)dx$$
For $a>1$ we have $$I =\lim_{n\rightarrow \infty}n^{\frac{1}{b}}\cdot 0$$
How do i solve it Help me please
Let $u=n^{a/b}x$, we have $$ n^{a/b}\int_0^1\frac{f(x)}{1+n^ax^b}dx=\int_0^{n^{a/b}}\frac{f(n^{-a/b}u)}{1+u^b}du=\int_0^{+\infty}g_n(u)du $$ with $g_n(u)=\frac{f(n^{-a/b}u)}{1+u^b}\chi_{[0,n^{a/b}]}(u)$. $(g_n)$ converges pointwise toward $g(u)=\frac{f(0)}{1+u^b}$ and $|g_n(u)|\leqslant\frac{\|f\|_{\infty,[0,1]}}{1+u^b}\in\mathcal{L}^1(\mathbb{R}^+)$ because $b>1$ (I suppose $f$ continuous). Using the dominated convergence theorem we have $$ \lim\limits_{n\rightarrow+\infty}n^{a/b}\int_0^1\frac{f(x)}{1+n^ax^b}dx=f(0)\int_0^{+\infty}\frac{du}{1+u^b} $$
