Hi I know the following sequence converges pointwise to $t$ from the limit but I am not sure if it converges uniformly for $t \in [0,1]$ and $n≥1$ $$f_n(t)= \frac {nt}{n+t}$$
I have found $|f_n(t)-f(t)|$ to be $$\frac {-t^2}{n+t}$$ But I am not sure how to get to uniform convergence from here. I believe it converges to $t$ uniformly, but could someone please clarify?
First of all, a minor mistake, your calculation $$|f_n(t)-f(t)|=\frac{-t^2}{n+t}$$ is wrong. For example, for $n=1,t=1$, the left side equals $$\left|\frac{1\cdot 1}{1+1} - 1\right| = \left|\frac12\right|=\frac12$$ while the right side equals $$\frac{-1^2}{1+1}=-\frac12$$ Similar, but not the same :)
You are almost there. Remember, convergence is uniform if, for all $\epsilon$, there exists some $N$ such that $\sup_{t\in[0,1]} |f_n(t)-f(t)| < \epsilon$ for all $n>N$.
In your case, the supremum becomes a maximum, and you already (almost) calculated $|f_n(t)-f(t)|$. The fact that $$\frac{t^2}{n+t} \leq \frac{1}{n+t}\leq \frac1n$$ should help you through the final steps.