# How to calculate $\displaystyle\lim_{x \to \infty} \left ( \frac{x+2}{x} \right )^{x}=e^{2}$

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?

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Take logarithms first, and see if you can massage it into something more amenable to further manipulation. –  Guess who it is. Nov 29 '10 at 7:03
How do you define $e$? –  AD. Nov 29 '10 at 8:53

$\lim _{x\rightarrow \infty }\left( \dfrac {x+2} {x}\right) ^{x}$ =$\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$
and it's easy for $\lim _{x\rightarrow \infty }\left( \dfrac {x} {x}+\dfrac {2} {x}\right) ^{x}$=$\lim _{x\rightarrow \infty }\left( 1+\dfrac {2} {x}\right) ^{x}$
$\lim _{x\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim _{\dfrac {x} {2}\rightarrow \infty }\left[ \left( 1+\dfrac {2} {x}\right) ^{\dfrac {x} {2}}\right] ^{2}$=$\lim {t\rightarrow \infty }\left[ \left( 1+\dfrac {1} {t}\right) ^{t}\right] ^{2}$($\dfrac {x} {2}=t$)=$e^{2}$(use $\lim {t\rightarrow \infty }$$\left( 1+\dfrac {1} {t}\right) ^{t}=e) - Find a step-by-step derivation here (and not only for this special problem): http://www.wolframalpha.com/input/?i=limit+((x%2B2)/x)%5Ex+x-%3Eoo (if the link doesn't work copy and paste the whole line) ...or go to http://www.wolframalpha.com directly and type: "limit ((x+2)/x)^x x->oo" Click on "Show steps" - Done! ;-) - Observe that$$\left( \dfrac{x+2}{x}\right) ^{x}=\left( \left( 1+\dfrac{1}{\frac{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*}$$For instance, by the same argument, for k\in\mathbb{N}:$$\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$e^2$. - 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. - That's a weird definition of$e$. It's certainly a true property, but weird to use that as a definition... – F'x Nov 29 '10 at 19:27 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. –  Vladimir Sotirov Nov 29 '10 at 20:07