Can we cancel out $1$ after replacing $e^x$ in $e^x-1$ by its Taylor expansion when calculating a limit?

Question

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!

Answer 1

Hint: Substitute $$e^x-1=t$$ in your Limit-function.

Answer 2

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 $.

Answer 3

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 $e^0=1$.

Answer 4

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 $$\lim_{x\rightarrow0}\frac{e^x-1}{x}=\lim_{t\rightarrow0}\frac{t}{\ln(1+t)}=\frac{1}{\lim\limits_{t\rightarrow0}\frac{\ln(1+t)}{t}}=1.$$