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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?

  • 2
    How do you define $e$?2010-11-29

6 Answers 6

16

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.

  • 0
    @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.2010-11-30
10

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

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

6

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 $

2

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

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$)

1

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! ;-)