Question

I know that Euler's number e is defined as the limit of (1+1/x)^x as x approaches infinity.

Answer 1

I would like to know how to solve this limit (without the use of calculus). I've done part of it.

Answer 2

\[\lim_{x \rightarrow \infty} (1+\frac{ 4.8 }{ x })^x\]

Answer 3

I know the answer is e^(4.8)

Answer 4

This is my work so far: \[\ln((1+\frac{ 4.8 }{ x })^x) = x \times \ln(1+\frac{ 4.8 }{ x})\]
You could then use one of the limit laws that says that the limit of the products is the product of the limits. \[\lim_{x \rightarrow \infty} x \times \lim_{x \rightarrow \infty} \ln(1+\frac{ 4.8 }{ x })\] The the first limit is infinity multiplied by the second is ln(1) =0. I took the limit inside because of continuity.

Answer 5

Then you would see \[\infty \times 0\] which I feel would qualify as an indeterminate form. That's where I don't know how to proceed.

Answer 6

we can use Lhopitals rule now

Answer 7

but first you need to make an indeterminate form of
\( \large \frac {\infty } {\infty} \) or \( \large \frac {0 } {0} \)

Answer 8

Hmm, yeah that's where I got confused. I'm not quite sure how to change it into that form, unless you could divide one limit by the other ?

Answer 9

In which case that would be infinity divided by 0 or vise versa, which isn't what we want.

Answer 10

$$\Large {\lim_{x \to \infty }\ln((1+\frac{ 4.8 }{ x })^x)
\\ \lim_{x \to \infty }= x \cdot \ln(1+\frac{ 4.8 }{ x})
\\ \lim_{x \to \infty }= \frac{ \ln(1+\frac{ 4.8 }{ x})}{\frac1x}

}$$

Answer 11

because x = 1 / (1/ x )

Answer 12

$$ \Large { x= \frac{1}{\frac1x }} $$

Answer 13

and the limit would be of the indeterminate form infinity over infinity? So we could then use l'hospitals rule and take the derivative of the top and bottom?

Answer 14

it would be indeterminate form of 0/0

Answer 15

$$



\Large {\lim_{x \to \infty }\ln((1+\frac{ 4.8 }{ x })^x)
\\ \lim_{x \to \infty }= x \cdot \ln(1+\frac{ 4.8 }{ x})
\\ \lim_{x \to \infty }= \frac{ \ln(1+\frac{ 4.8 }{ x})}{\frac1x}\\ ~\\
=\frac{ \ln(1+\frac{ 4.8 }{ \infty})}{\frac{1}{\infty}} = \frac{\ln(1 + 0)}{0}= \frac{\ln(1)}{0}= \frac{0}{0}
}


$$

Answer 16

Alright, I'm following so far.

Answer 17

and therefore we can use Lhopitals rule :)

Answer 18

alright, thanks a lot :)

Answer 19

did you do Lhopitals rule?

Answer 20

Yes, sorry got sidetracked lol. I believe I got ln * 4.8

Answer 21

$$
\Large {\lim_{x \to \infty }\ln((1+\frac{ 4.8 }{ x })^x)
\\ \lim_{x \to \infty }= x \cdot \ln(1+\frac{ 4.8 }{ x})
\\ \lim_{x \to \infty }= \frac{ \ln(1+\frac{ 4.8 }{ x})}{\frac1x}
\\ \huge \Rightarrow^{Lhopitals}\\
\LARGE \lim_{x \to \infty } \frac{ \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{x^2} }{-\frac{1}{x^2}}
\\ = \lim_{x \to \infty } \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{x^2} \cdot\frac{-x^2}{1}

\\ = \lim_{x \to \infty } \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{\cancel{x^2}} \cdot\frac{-\cancel{x^2}}{1}


}


$$

Answer 22

Oh yeah, chain rule.

Answer 23

Thanks so much @perl I appreciate how detailed your explanation was. It was very helpful.

Answer 24

$$

\Large \rm {
Suppose~ that~ the~ limit ~ exists,~ call~ it ~ L
\\
L = \lim_{x \rightarrow \infty} (1+\frac{ 4.8 }{ x })^x
\\~\\ Then ~ it~ follows~
\\ \ln(L) = \ln \left( \lim_{x \rightarrow \infty} (1+\frac{ 4.8 }{ x })^x \right)
\\=\lim_{x \to \infty }\ln(1+\frac{ 4.8 }{ x })^x
\\ =\lim_{x \to \infty } x \cdot \ln(1+\frac{ 4.8 }{ x})
\\ =\lim_{x \to \infty } \frac{ \ln(1+\frac{ 4.8 }{ x})}{\frac1x}
\\ \huge \Rightarrow^{Lhopitals}\\
\LARGE =\lim_{x \to \infty } \frac{ \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{x^2} }{-\frac{1}{x^2}}
\\ = \lim_{x \to \infty } \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{x^2} \cdot\frac{-x^2}{1}

\\ = \lim_{x \to \infty } \frac{1 }{(1+\frac{ 4.8 }{ x}) } \cdot \frac{-4.8}{\cancel{x^2}} \cdot\frac{-\cancel{x^2}}{1}

\\ = \lim_{x \to \infty } \frac{4.8 }{(1+\frac{ 4.8 }{ x}) }
\\~\\ = \frac{4.8}{(1+\frac{ 4.8 }{ \infty }) } = \frac{4.8}{(1+0 )}= \frac{4.8}{1}
=4.8
\\~\\ \therefore \\~\\
\\ \ln(L) = 4.8 ~ \iff L = e^{4.8}
}

$$

Answer 25

your welcome :)

Answer 26

:)