Convergence of sequence of functions values at a sequence If $\{h_{k}\}_{k=1}^{\infty}$ is a sequence of real-valued continuous functions on the real line, and $0<h_{k}(x)\leq 1,\;\forall x\in\mathbb{R}$ and all $k$. Assume that $h_{k}(x)\to 0$ uniformly on $\mathbb{R}$. Is it true that $h_{k}(x+x_{k})\to 0$ as $k\to \infty$, for any sequence of points $\{x_{k}\}\subset \mathbb{R} $ and any $x\in \mathbb{R}$? If so how to prove it?
Thanks for help in advance!
Ok, what I have tried:
I'm trying to use induction on $x_{k}$: Fix any $x\in\mathbb{R}, $we have 
for $x_{1}$: $h_{k}(x+x_{1})\to 0$
for $x_{2}$: $h_{k}(x+x_{2})\to 0$
for $x_{3}$: $h_{k}(x+x_{3})\to 0$
.
.
.
for $x_{m}$: $h_{k}(x+x_{m})\to 0$
so, its true for all $m$, hence for $k$. Not sure!
Edit: I had a typo, sorry. My previous question was about $h_{k}(x_{k})\to 0$, but the correct question is about $h_{k}(x+x_{k})\to 0$
 A: This should be fairly direct.
The statement that $h_k(x) \to 0$ uniformly on $\mathbb{R}$ means that for any $\epsilon > 0$, there exists some $K_\epsilon$ such that $k>K_\epsilon$ implies that $|h_k(x)|<\epsilon$ for all $x \in \mathbb{R}$. If $\{x_k\}$ is some sequence in $\mathbb{R}$, then we certainly have that $|h_k(x_k)| < \epsilon$ for all $k>K_\epsilon$, and so $h_k(x_k) \to 0$.
Uniform convergence is a very powerful property.
A: The idea Greg Zitelli points out in his answer should work with the new version of the problem too.
I'd just like to add that your idea of proof (if I understand it correctly) won't work. Here's an example that should demonstrate what the problem is. Let $$f_n(x)=\begin{cases}0;&x<n\\1;&\text{otherwise}\end{cases}$$
and let $x_n=n$. Then for every term $x_i$ of this sequence you have: $f_n(x_i)\to0$, but $f_n(x_n)\to1$. Of course here the convergence is not uniform and the functions are not continuous, but it still shows that the convergence of $f_n(x_n)$ does not follow from that of $f_n(x_i)$.
