Let $f_n(x)={x^n\over n}+1$. I need to find the domain of it's uniform convergence. I tried the $M_n$-test. So, I got the point limit $f(x)=1$ when $x\in [0,1]$. For $x>1$ the point limit wont exist, hence no question of uniform convergence arises. Thus, the convergence in uniform in $[0,1]$. That goes right now hopefully.
UPDATE Also, now I observed the convergence will hold uniformly in $[-1,0]$ by similar argument. For $\vert x\vert >1$, the point limit won't exist. So, the domain of uniform convergence should be $[-1,1]$
That is not true. I'll only discuss the sequence for nonnegative $x$. The negative $x$ I'll leave it for you. For $x>1$ the limit is infinity (why?). So $f(x) =1$ cannot be the limit function for ALL $x$! For $x=1$ the limit is $1$. For $x\in[0,1)$ we have $f(x) =1$. That was all pointwise limits.
Now let's check uniform convergence where there is pointwise convergence so only on $[0,1]$. So $f(x) =1$ and \begin{align} \sup_{x\in[0,1]} |f_n(x) - f(x) |\leq \frac{1} {n} \to 0 \end{align} So uniform convergence on $[0,1]$. Now do the negative half line yourself and conclude.