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?
 A: 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 $0$.

A: One may observe that, as $u \to 0$, by the Taylor expansion we have
$$
e^u=1+u+O(u^2)
$$ giving, as $n \to \infty$,
$$
n\left(e^{x/n}-1\right)=n \times \frac{x}n+O\left(\frac{x}{n}\right)
$$ and
$$
n\left(e^{x/n}-1\right) \to x
$$
A: $$\lim_{n \rightarrow \infty} n(e^{x/n}-1) = \lim_{y \rightarrow 0} \frac{x(e^y-1)}{y} = \lim_{y \rightarrow 0}xe^y = x$$
This requires L'Hopitals rule and a substitution $y = \frac{x}{n}$
A: $$ \lim_{n \rightarrow \infty} n(e^{x/n}-1)=\lim_{n \rightarrow \infty} \frac{(e^{x/n}-1)}{\frac{1}{n}}$$
using L'Hopital rule
$$\lim_{n \rightarrow \infty} \frac{(-\frac{x}{n^2}e^{x/n})}{-\frac{1}{n^2}}=x$$
A: One way to approach this is to replace $n \rightarrow \infty$ with $h = \frac{1}{n} \rightarrow 0$ and then observing that 
$$\lim_{h\rightarrow 0} \frac{e^{hx} - 1}{h}$$ is the derivative of a function $f(y) = (e^x)^y$ evaluated at $0$ since
$$ f'(y) = \lim_{h\rightarrow0} \frac{(e^x)^h - e^y}{h} $$ and of course $e^0 = 1$. $$f'(0) = \ln(e^x) (e^x)^0 = x$$
A: 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
