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

Question

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?

Answer 1

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.

Answer 2

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

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

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

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

Answer 3

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

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

Answer 4

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

Answer 5

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$)

Answer 6

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! ;-)