$\displaystyle f_{n}(x)=log(x+\frac{1}{n})
f_{n}(x)=cos(\frac{x}{n}) $
on (0,1)
I'm a little confused about what you have written.
I assume those are two different functions?
$\displaystyle f_n(x)=\log(x+\frac{1}{n})$ and $\displaystyle f_n(x)=\cos(\frac{x}{n})$
Also, are you using $\displaystyle \log(x)$ to mean the same thing as $\displaystyle \ln(x)$?
Second function: We know that this function converges to $\displaystyle \cos(0)=1$, the question is whether it does so uniformly. Using the difference-to-product identity, we have:
$\displaystyle \left|\cos(0)-\cos\left(\frac{x}{n}\right)\right| = \left|-2\sin\left(\frac{x}{2n}\right)\sin\left(-\frac{x}{2n}\right)\right|\leq 2\left|\sin\left(\frac{x}{2n}\right)\right|$ because $\displaystyle |\sin(y)|\leq 1$ for all $\displaystyle y$.
Since $\displaystyle 2\sin\left(\frac{x}{2n}\right)$ is increasing and positive on $\displaystyle (0,1)$, we know that $\displaystyle 2\left|\sin\left(\frac{x}{2n}\right)\right|\leq 2\sin\left(\frac{1}{2n}\right)$.
It is now clear that for all $\displaystyle \epsilon>0$, we can choose $\displaystyle n$ large enough so that $\displaystyle \left|1-\cos\left(\frac{x}{n}\right)\right| \leq 2\sin\left(\frac{1}{2n}\right)<\epsilon$
If you need an explicit formula, I think choosing $\displaystyle n>\frac{1}{2\sin^{-1}(\frac{\epsilon}{2})}$ will do the trick.