Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity 
Let $f: (-\infty,0]\rightarrow\mathbb R$ which is a function: $f(t)=\arcsin(e^{t})$ for $t\le0$.Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity

My try:
For $t\neq0$: $$f'(t)=\frac{e^{t}}{\sqrt(1-e^{2t})}$$$$f'(t)>1$$However $f'(t)$ has no upper limit so function $f'$ is not limited.We know that limited derivative $ \Leftrightarrow $ Lipschitz continuity for differentiable functions. So the first observation seems to be that $ f $ does not meet the Lipschitz continuity. But $f$ it is not differentiable in the whole domain so I don't know if I could use this observation.Moreover I know that function is uniform continuity when $$\forall_{\epsilon>0}\exists_{\delta>0} \forall_{x \in D}\forall_{y \in D}(|x-y|< \delta\Rightarrow |f(x)-f(y)|<\epsilon)$$Unfortunately I don't have idea how to show that $f$ (no)meets this condition.Can you get me some tips with it?
 A: For the first part, note that if $f$ were Lipschitz continuous on $(-\infty,0]$ with constant $L$, then the derivative $f'$, where it exists, would satisfy $|f'(x)|\leq L$. But this does not hold since $|f'(t)|\rightarrow +\infty$ for $t\rightarrow 0^-$.
Another argument to prove that $f$ is not Lipschitz continuous on $(-\infty,0]$ resorts to the definition. If $f$ were Lipschitz continuous on $(-\infty,0]$, then there would be a constant $L>0$ such that $|f(x)-f(y)|\leq L|x-y|$ for all $x,y\in (\infty,0]$. Take $x=0$ and $y=-\frac 1n$: the previous conditions would read
$$
\forall n\in \mathbb{N}\quad|1-\arcsin e^{-\frac 1n}|\leq L\cdot \frac 1n,
$$
that is
$$
n|1-\arcsin e^{-\frac 1n}|\leq L,
$$
which is not possible to satisfy for all $n\in \mathbb{N}$.

As for the second part, you have to use some properties of uniformly continuous function, since the definition is not so helpful in these cases. Here, $f$ is continuous over $(-\infty,0]$ and $\lim_{x\rightarrow -\infty} f(x)$ exists, in particular it is equal to $0$. This implies uniform continuity over $(-\infty,0]$.
A: You have the derivative wrong: $f'(t)= e^t/\sqrt {1-e^{2t}}.$ You can use your Lipschitz-derivative criterion by noting that if $f$ were Lipschitz on $(-\infty,0],$ it would be Lipschitz on $(-\infty,0).$ On the latter interval $f$ is differentiable.
