Can we cancel out $1$ after replacing $e^x$ in $e^x-1$ by its Taylor expansion when calculating a limit? So I am to prove the following formula

$$\lim_{x \to 0} \frac{e^x -1}{x} =1$$

Now, I write,
$$
\begin{split}
\lim_{x \to 0} \frac{e^x -1}{x} 
   &= \lim_{x \to 0} \frac{(1+x+x^2/2!+x^3/3!+...) -1}{x} \\
   &= \lim_{x \to 0} \frac{x+x^2/2!+x^3/3!+...}{x} \\
   &= \lim_{x \to 0}1+x/2 +x^2/6+...\\
   &= 1
\end{split}
$$
Is it permissible to cancel out the $1$ at the numerator  and the calculation after that? Here we are dealing with an infinite series expansion of $e^x$ ,so I am very much confused about this method.
By the way,I am not here for any other rigorous proof; I just want to know if my method is correct or not and why it is so.
Thanks in advance!
 A: I'd suggest noticing that the limit you want to evaluate is 
$$
\lim_{x\to0}\frac{e^x-e^0}{x-0},
$$
which is the definition of the derivative at $0$ of the exponential function. So if you know that the derivative of $e^x$ is $e^x$, then you immediately get the answer $e^0=1$.
A: We know that $$\lim_{t\rightarrow0}(1+t)^{\frac{1}{t}}=e.$$
Thus, since $\ln$ is a continuous function, we obtain:
$$\lim_{t\rightarrow0}\frac{\ln(1+t)}{t}=\lim_{t\rightarrow0}\ln(1+t)^{\frac{1}{t}}=\ln\lim_{t\rightarrow0}(1+t)^{\frac{1}{t}}=\ln{e}=1.$$
Now, let $\ln(1+t)=x$.
Thus, $t=e^x-1$ and $$\lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{t\rightarrow0}\frac{t}{\ln(1+t)}=\frac{1}{\lim\limits_{t\rightarrow0}\frac{\ln(1+t)}{t}}=1.$$
A: Hint: Substitute $$e^x-1=t$$ in your Limit-function.
A: Since
$1+x \le e^x
\le \dfrac1{1-x}
$
for
$0 \le x \lt 1$
(compare terms in the power series
or see below),
and
$\dfrac1{1-x}
\le 1+x+2x^2
$
for
$0 \le x \le \frac12
$,
$1 
\le \dfrac{e^x-1}{x}
\le 1+2x
$
for
$0 \le x \le \frac12
$.
To show
$e^x
\le \dfrac1{1-x}
$,
if
$f(x)
=(1-x)e^x
$,
$f(0) = 1$
and
$f'(x)
=(1-x)e^x-e^x
=-xe^x
\le 0
$
for
$0 \le x \le 1
$
so
$f(x)
=f(0)+\int_0^x f'(t) dt
\le 1
$.
