I'm looking over uniform convergence in my textbook and want to make sure I'm doing the problems correctly. The question is
Determine the pointwise limit of $(f_n)$ on the indicated interval, and state whether the convergence is uniform.
a) $f_n(x)=\sqrt[n]{x}$, on $[0,1]$
b) $f_n(x)= \frac{e^x}{n^2}$ on $(1, \infty)$
c) $f_n(x)=x^n-x^{2n}$ on $[0,1]$
d) $f_n(x)=\frac{2nx}{1+n+3x}$ on $[0,\infty)$.
What I have so far is
a) $f(x)=\begin{cases} 0 &\text{if}\, x=0\\ 1 &\text{if}\, x\in(0,1]\ \end{cases}$ and that $f_n(x)$ does not converge uniformly. Let $\varepsilon=\frac{1}{2}$. Choose $x_n=e^{-n}$, then $\left|f(x_n)-f_n(x_n)\right|=\left|1-\sqrt[n]{e^{-n}}\right|=1-\frac{1}{e}>\frac{1}{2}\ge\varepsilon$. I believe you could also argue that since $f(x)$ is discontinuous and each $f_n(x)$ is continuous, then $f_n(x)$ cannot converge uniformly since uniform convergence preserves continuity.
b) $f(x)=0$ and that $f_n(x)$ does not converge uniformly. Let $\varepsilon=1$. Choose $x_n=\ln(n^2)$. Then $\left|f_n(x_n)-f(x_n)\right|=\left|\frac{e^{\ln(n^2)}}{n^2}-0\right|=1\ge \varepsilon$.
c) $f(x)=0$ and that $f_n(x)$ does not converge uniformly. Let $\varepsilon=\frac{1}{4}$. Choose $x_n=\sqrt[n]\frac{1}{2}$. Then $\left|f_n(x_n)-f(x_n)\right|=\left|\left(\sqrt[n]\frac{1}{2}\right)^n-\left(\sqrt[n]\frac{1}{2}\right)^{2n}-0\right|=\frac{1}{2}-(\frac{1}{2})^2=\frac{1}{4}\ge \varepsilon$.
d) $f(x)=2x$ and that $f_n(x)$ does not converge uniformly. Let $\varepsilon=1$. Choose $x_n=n$. Then $\left|f_n(x_n)-f(x_n)\right|=\left|\frac{2nn}{1+n+3n}-2n\right|=2n\left|\frac{n}{1+4n}-1\right|=2n\left|\frac{1+3n}{1+4n}\right|\ge2n\frac{1}{2}=n\ge1=\varepsilon$.