2
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Yes it is permissible because you are just looking at the series of $e^x-1$. $\endgroup$ – Anurag A Jul 15 at 18:54
  • 2
    $\begingroup$ 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. $\endgroup$ – WaveX Jul 15 at 18:55
  • 1
    $\begingroup$ 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. $\endgroup$ – user247327 Jul 15 at 19:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Andreas Blass Jul 15 at 19:25
  • 1
    $\begingroup$ 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.) $\endgroup$ – Andreas Blass Jul 15 at 19:39
8
$\begingroup$

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 $e^0=1$.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.