Finding $\lim\limits_{n \rightarrow \infty} n(e^{x/n}-1)$ Since $\lim\limits_{n \rightarrow \infty} (1+x/n)^n = e^x$

I should be able to find $\lim\limits_{n \rightarrow \infty} n(e^{x/n}-1)$ in some related way, but I have been struggling.

Can anyone provide any insight?
Another approach (if you are not familiar with Taylor expansions): setting $h=\frac{1}{n} \xrightarrow[n\to\infty]{} 0$, you can rewrite
$$
n(e^{\frac{x}{n}} -1) = \frac{e^{h\cdot x}-1}{h} = \frac{e^{h\cdot x}-e^{0\cdot x}}{h-0}.
$$
Does that expression look familiar?

Further hint:


   Consider the function $g(t) = e^{tx}$, which is differentiable at $0$.
lim_{x->∞} n(e^{x/n}-1)=x: For we need to find lim_{y->∞}((e^{x/y}-1)/(1/y))=lim_{y->_{∞}}((e^{x/y}.(-x/y²))/(-1/y²))=x