A sequence of continuous functions on $[0,\infty)$ converging uniformly such that each $f_n$ has a zero but the limit function is nowhere zero. 
We have a sequence of continuous functions $f_n : [0,\infty) \to \Bbb R$ such that $f_n \to f$ uniformly. Also each $f_n$ has a zero. But $f$ does not have any zero. Then what is the example of such a sequence of functions and their limit function.

I am trying to think about a sequence of functions such that their limit function is $f(x)=1 \quad \forall x \in [0,\infty)$. However, I am really stuck on this for a long time and completely clueless after trying many functions $f_n$.
Here is the link where I proved (after getting help) that if the domain is $[a,b]$ then $f$ will always have a zero because of sequential compactness of $[a,b]$. In case of $[0,\infty)$ it is not possible. 
 A: Consider $$f_n(x)=\frac{1}{x+1}-\dfrac{\sin x}{n}.$$ Note that every $f_n$ has at least a zero. But, its limit $f(x)=\dfrac{1}{x+1}$ has not zeroes.
$$f_n(x)=0\iff h(x)=(x+1)\sin x-n=0.$$ But $h(0)=-n$ and $h\left(\dfrac{(4n+1)\pi}{2}\right)=\dfrac{(4n+1)\pi}{2}+1-n>0.$ Since $h$ is continuous it must have a zero in the interval $\left(0,\dfrac{(4n+1)\pi}{2}\right).$
Finally, note that $$|f_n(x)-f(x)|=\dfrac{|\sin x|}{n}\le \dfrac 1n,$$ from where we can conclude that $f_n\to f$ uniformly.
A: I'll mention a whole class of such functions, without going into details. Let
$$g_n(x) = \frac{(x-n)^2}{1+(x-n)^2}.$$
Now let $f$ be any positive continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}f(x) = 0.$ Then $g_n f \to f$ uniformly on $[0,\infty).$ Because $(g_nf)(n) = 0,$ this gives many examples for the problem at hand.
I can go into this further if there are questions.
A: Hint: Use $f(x) =\frac{1}{1+x}$.
A: Take $f(x) = e^{-x}$, and let $f_n(x)$ equal $f(x)$ for $x \notin[n,n+1]$, but
$f_n(x)$ takes the value computed by linearly interpolating the points
$(n,f(n)), (n+{1 \over 2}, 0), (n+1, f(n+1))$. Then
$\|f-f_n\|_\infty = f(n+{1 \over 2})$ hence converges uniformly, and $f_n(n+{1 \over 2}) = 0$. Since $f(x) \neq 0$, we are finished.
