For what values of $k$ is the function uniformly convergent? Given that $k\geq 0$ and $f_n:\mathbb{R^+}\rightarrow \mathbb{R^+}$ defined by $f_n:=\frac{x^k}{x^2+n}$. For what values of $k$ is $\{f_n\}$ uniformly convergent.
 A: I think I have gotten the answer. The sequence converges pointwise to $0$ for all $k$. But for $k>2$, taking $x_n=\sqrt n$, we have $f_n(x_n)=\frac{n^\frac{k}{2}}{2n}\rightarrow +\infty$. So, sup over $x\in R$ of $|f_n(x)|=+\infty$. Also, for $k=1$, using the same sequence gives $f_n(x_n)\rightarrow \frac{1}{2}$. Therefore sup over $R$ of $|f_n(x)|\geq \frac{1}{2}$.   So the sequence is not uniformly convergent for $k\geq 2$. For $0\leq k<2$, $|f_n(x)|<\min\{\frac{x^k}{x^2}, \frac{x^k}{n}\}$. Taking sup over $x\in R$ on both sides gives that $f_n$ is uniformly convegrent. 
A: A local maximum value of $f_n(x)$ occurs when $f_n'(x)=0$.  
Setting the derivative of $f_n$ to zero reveals that 
$$f_n'(x)=\frac{kx^{k-1}(x^2+n)-2x^{k+1}}{(x^2+n)^2}=0$$
Thus, the maximum occurs at $x^2=\frac{kn}{2-k}$.  Obviously, a local maximum exists only when $k<2$.  For $k<2$, we have 
$$f_n(x) \le \frac{\left(\frac{kn}{2-k}\right)^{k/2}}{\frac{kn}{2-k}+n}=O\left(n^{k/2-1}\right)\to 0$$
Thus, $f_n$ converges uniformly to $0$ for all $k<2$.
For $k\ge 2$, we may choose $x$ so large that $x^k>\frac12(x^2+n)$ (e.g., take $x^k=n+1$) and thus the convergence of $f_n$ is not uniform.
