I know from an online calculator http://www.numberempire.com/derivatives.php that $\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$. How do you calculate this step by step?

  • $\begingroup$ Take logarithms first, and see if you can massage it into something more amenable to further manipulation. $\endgroup$ – J. M. is a poor mathematician Nov 29 '10 at 7:03
  • 2
    $\begingroup$ How do you define $e$? $\endgroup$ – AD. Nov 29 '10 at 8:53

This is more or less by definition of $e$, depending on which definition you use. Do you know the definition $e=\lim_{n\rightarrow\infty} (1+\frac{1}{n})^n$? Then set $x=2n$ and you are done.

  • $\begingroup$ That's a weird definition of $e$. It's certainly a true property, but weird to use that as a definition... $\endgroup$ – F'x Nov 29 '10 at 19:27
  • 12
    $\begingroup$ There's nothing weird about that definition of $e$. The quantity $(1+\frac1n)^n$ represents the $100%$ interest compounded $n$ times over the year. Taking the limit as $n$ goes to infinity gives you how much money you earn after continuous compounding. Then $e^x$ represents the amount of money you earn if the interest is $100x%$. It is certainly plausible from this definition that the derivative of $e^x$ is $e^x$ and in any case one certainly gets the same power series by treating $\lim_{n\to\infty}(1+\frac xn)^n$ as a limit of power series. $\endgroup$ – Vladimir Sotirov Nov 29 '10 at 20:07
  • $\begingroup$ @FX_, you are free to use whatever definition you please. The way you have worded your comment, it reads like a rant, so I am not going to say anything further. But if you want to know, how to arrive at this definition naturally (e.g. the way Vladimir Sotirov has described, but there are also other ways), feel free to post a question. $\endgroup$ – Alex B. Nov 30 '10 at 1:58

Observe that

$$\left( \dfrac{x+2}{x}\right) ^{x}=\left( \left( 1+\dfrac{1}{x/2} \right) ^{x/2}\right) ^{2}.$$

Hence, by the definition of $e$

$$e=\lim_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$

we have

$$\begin{eqnarray*} \lim_{x\rightarrow \infty }\left( \dfrac{x+2}{x}\right) ^{x} &=&\lim_{x\rightarrow \infty }\left( \left( 1+\frac{1}{x/2}\right) ^{x/2}\right) ^{2} \\ &=&\left( \lim_{x\rightarrow \infty }\left( 1+\frac{1}{x/2}\right) ^{% x/2}\right) ^{2} \\ &=&\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right) ^{2} \\ &=&e^{2} \end{eqnarray*}$$

For instance, by the same argument, for $k\in\mathbb{N}$, we deduce that

$$\lim_{x\rightarrow \infty }\left( \dfrac{x+k}{x}\right) ^{x}=\left( \lim_{y\rightarrow \infty }\left( 1+\frac{1}{y}\right) ^{y}\right) ^{k}=e^{k}$$


For ease rewrite as

$$ \lim_{n\rightarrow \infty}\left(1+\frac{2}{x}\right)^x. $$

First we compute

$$ \lim_{n\to\infty}\ln\left(\left(1+\frac{2}{x}\right)^x\right). $$

Using laws of logarithms we get

$$ \lim_{n\to\infty}x\ln\left(1+\frac{2}{x}\right) =\lim_{n\to\infty}\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}. $$

We are now in a position to apply L'Hopital's Rule. Taking derivatives gives

$$ \lim_{n\to\infty} \frac{\left(\frac{1}{1+\frac{2}{x}}\right)\left(-\frac{2}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{n\to\infty}\frac{2}{1+\frac{2}{x}}=2. $$

Now, your limit is

$$ \lim_{n\to\infty}e^{\ln\left(\left(1+\frac{2}{x}\right)^x\right)}=e^2 $$


Rewriting the function whose limit is taken $$\left(1 + \frac{2}{x}\right)^x,$$ we immediately recognize the definition of $e^2$.


It's easy to obtain

$\lim _{x\rightarrow \infty }\left( \dfrac {x+2} {x}\right) ^{x}$ =$\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$

and it's easy for $\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$=$\lim _{x\rightarrow \infty }\left( 1+\dfrac {2} {x}\right) ^{x}$

do some conversion

$\lim _{x\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim _{\dfrac {x} {2}\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim {t\rightarrow \infty }\left[ \left( 1+\dfrac {1} {t}\right) ^{t}\right] ^{2}$($\dfrac {x} {2}=t$)=$e^{2}$(use $\lim {t\rightarrow \infty }$$\left( 1+\dfrac {1} {t}\right) ^{t}$=$e$)


Find a step-by-step derivation here (and not only for this special problem): http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x)%5Ex+x-%3Eoo

(if the link doesn't work copy and paste the whole line)

...or go to http://www.wolframalpha.com directly and type:
"limit ((x+2)/x)^x x->oo"

Click on "Show steps" - Done! ;-)


Your Answer

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