# 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!

• Yes it is permissible because you are just looking at the series of $e^x-1$. – Anurag A Jul 15 at 18:54
• Sure. Think about it as $(-1)+1+x+\dots$ in the numerator and perhaps it easier to see that what you are doing is allowed. All of the (countably) infinite terms remaining in the numerator can certainly have an $x$ factored out so the $x$ in the denominator can be cancelled out. – WaveX Jul 15 at 18:55
• How do you know that any series converges? Do you know the "ratio test"? A general term in that is $a_n= \frac{x^{n-1}}{n!}$. So $\frac{a_{n+1}}{a_n}= \frac{x^n}{(n+1)!}\frac{n!}{x^{n-1}}= \frac{1}{n+1}x$. That goes to 0 as n goes to infinity for all x so that series converges for all x. – user247327 Jul 15 at 19:18
• In addition to convergence of $1+(x/2)+(x^2/6)+\dots$, you need to know that this sum approaches $1$ as $x$ tends to $0$. That's not obvious unless you have some theorems about power series available. – Andreas Blass Jul 15 at 19:25
• I'd use the theorem that the functions defined by convergent power series within the interval of convergence are continuous. (That it seems trivial is one reason for my comment --- to point out that it's not trivial and needs proof.) – Andreas Blass Jul 15 at 19:39

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 $$e^0=1$$.

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

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

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 $$f(x) =f(0)+\int_0^x f'(t) dt \le 1$$.