Is $f$ uniformly continuous Prove/Disprove : Let $f:(0,1)→\mathbb{R}$ be Continuous. The condition $f(1/n)$$\rightarrow$$1/2$ and $f(1/n^2)$$\rightarrow$$1/4$  imply that $f $ is uniformly continuous.
The condition implies everything! This because it is not true.

($f\left(\frac{1}{n}\right)\rightarrow2$ implies: $f\left(\frac{1}{n^{2}}\right)\rightarrow2$
wich contradicts $f\left(\frac{1}{n^{2}}\right)\rightarrow4$)

$p\rightarrow q$ is true if $p$ is not true.

So indeed the condition implies that $f$ is uniformly continuous. There is no function that satisfies this condition and is not uniformly continuous. You cannot even find a function that satisfies this condition.
As pointed out in other answers, the conditions are self-contradictory; the sequence $f(1/n^2)$ is a subsequence of $f(1/n)$. If they have limits at all, they cannot have different ones.

It seems to me that the spirit of the question is to ask, 


  if $a_n,\ b_n\to 0$ but $\lim f(a_n)\neq \lim f(b_n)$, then can $f$ be uniformly continuous? 


Now my hint is: $f$ cannot extend to a continuous function $\hat{f}:[0,1]\to\mathbb{R}$. (Can you see why?) Why does this imply $f$ must not be uniformly continuous?