Proving that a function sequence converges uniformly to limit function I have the function sequence $f_n = n\sinh(x/n)$, $\forall n \in \mathbb{N}, \forall x \in [-1, 1]$ which I believe has the limit function $f = 1$ for $x = 1, f = 0$ for $x \in (-1,1)$ and $f = -1$ for $x = -1$.

How would I go about proving that $f_n$ converges to $f$ uniformly? I know that $f_n \rightarrow f$ iff $f_n \rightarrow f$ in the uniform topology. So I should somehow be able to use the metric that induces the uniform topology to prove that $f_n$ converges uniformly. Is it enough to show that the derivatives of the functions are bounded by some value not dependent on $x$?
$e^{t}=1+t+tg(t)$ where $g(t) \to 0$ as $t \to 0$ (and $g$ is continuous on $[0,1]$). [ To see this either use Taylor expansion to apply L'Hopital's Rule to $\frac {e^{t+-1-t}} t$]. 

Hence $n\frac {e^{x/n}-e^{-x/n}} 2 =x+\frac  x n [g(x/n)-g(-x/n)]$.

Note that $g(t)=\frac {e^{t}-1-t} t$ and $g(0)=0$ defines a continuous function on $[0,1]$ such that $e^{t}=1+t+tg(t)$ for all $t$. Being a continuous function $g$ is bounded Suppose $|g(t)| \leq M$ for all $t$. Then $|\frac  x n [g(x/n)-g(-x/n)]| \leq 2M \frac 1 n$ for all $x$. Hence $\sup_x  |\frac  x n [g(x/n)-g(-x/n)]|  \to 0$ proving uniform convergence.
$e^{t}=1+t+tg(t)$ where $g(t) \to 0$ as $t \to 0$ (and $g$ is continuous on $[0,1]$). [ To see this either use Taylor expansion to apply L'Hopital's Rule to $\frac {e^{t+-1-t}} t$]. 

Hence $n\frac {e^{x/n}-e^{-x/n}} 2 =x+\frac  x n [g(x/n)-g(-x/n)]$.

Note that $g(t)=\frac {e^{t}-1-t} t$ and $g(0)=0$ defines a continuous function on $[0,1]$ such that $e^{t}=1+t+tg(t)$ for all $t$. Being a continuous function $g$ is bounded Suppose $|g(t)| \leq M$ for all $t$. Then $|\frac  x n [g(x/n)-g(-x/n)]| \leq 2M \frac 1 n$ for all $x$. Hence $\sup_x  |\frac  x n [g(x/n)-g(-x/n)]|  \to 0$ proving uniform convergence.