Pointwise convergence but not uniformly convergence of $g_{n} \to 0$ when $g_{n}(x)=f\left(\frac{\sqrt[n]{x}}{1+ \sqrt[n]{x}}\right)$. Let $f:[0,1] \to \mathbb{R}$ a continuous function such $f(0)=f(\frac{1}{2})=0$ and $(f[0, \frac{1}{2}]) \not \subset \lbrace 0 \rbrace$.  For each $n \in \mathbb{N}$, let $g_{n}:[0, \infty ) \to \mathbb{R}$ defined for each $x \in [0, \infty)$ as $$g_{n}(x)=f\left(\frac{\sqrt[n]{x}}{1+ \sqrt[n]{x}}\right)$$
Prove the following:
(a) $\lbrace g_{n} \rbrace_{n=1}^{\infty}$ is pointwise convergent to $0$.
(b) $\lbrace g_{n} \rbrace_{n=1}^{\infty}$ is not uniformly continuous to $0$.
My attemp for (a) goes as follow: As
$$\left| \frac{\sqrt[n]{x}}{1+ \sqrt[n]{x}} \right|= \left| \left( \sqrt[n]{x}\right)  \frac{1}{1+ \sqrt[n]{x}} \right| = \left| \sqrt[n]{x} \right|   \left| \frac{1}{1+ \sqrt[n]{x}}  \right|$$. I want to use the fact that as $\sqrt[n]{x} \to 1$ as $n \to \infty$ for $x \in [0,1]$ and $\frac{1}{1+ \sqrt[n]{x}} \to \frac{1}{2}$ as $n \to \infty$ for $x \in [0,1]$ then we can bound 
$$\left| \frac{\sqrt[n]{x}}{1+ \sqrt[n]{x}} - \frac{1}{2} \right|< \epsilon.$$
For every $\epsilon>0$ and of course a particular $n\geq N(\epsilon) \in \mathbb{N}$. Then as $f$ is continuous in [0,1] then 
$$ g_{n}(x)=f\left(\frac{\sqrt[n]{x}}{1+ \sqrt[n]{x}}\right) \to f\left(\frac{1}{2}\right)=0$$
Im kind of lost in proving the not uniformly continuity of $g_{n} \to 0$. Any help ending the proof would be appreciated. Thanks
 A: We show that the function is not uniformly continuous even over $(1,\infty)$. First notice that since $f(x)$ is continuous over $[0,1]$ we have:$$0<x<\delta_\epsilon\quad,\quad |f(x)|<\epsilon $$therefore $$\forall n>N\quad\to \quad 0<{\sqrt[n]x\over 1+\sqrt[n]x}<\delta_\epsilon\to |g_n(x)|=|f({\sqrt[n]x\over 1+\sqrt[n]x})|<\epsilon$$also $0<{\sqrt[n]x\over 1+\sqrt[n]x}<\delta_\epsilon$ is equivalent to $$n>\left({1\over \delta_\epsilon}-1\right){1\over \ln x}=N$$. Uniform continuity requires that $N=\left({1\over \delta_\epsilon}-1\right){1\over \ln x}$ be a function only of $\epsilon$ while we see that a term $1\over\ln x$ makes it dependent also to $x$. Therefore $g_n(x)$ tends to $0$ though not uniformly.
A: For (a) you need to distinguish between $x=0$ and $x>0.$ If $x=0,$ then $g_n(x) = f(0)=0$ for all $n.$ If $x>0,$ then $x^{1/n} \to 1,$ hence $x^{1/n}/(1+x^{1/n}) \to 1/2,$ which implies $g_n(x) \to f(1/2)=0.$
For (b), we know that there is $c\in (0,1/2)$ such that $f(c)\ne 0.$ This $c$ can be written as $x/(1+x)$ for some $x\ge 0.$ We then have $g_n(x^n) = f(x/(1+x))=f(c)$ for all $n.$ Thus
$$\sup_{\mathbb [0,\infty)}|g_n| \ge |g_n(x^n)| = |f(c)| >0$$
for all $n.$ This shows $g_n$ does not converge uniformly to $0$ on $[0,\infty).$
