Question

question below

Answer 1

is \[\lim_{n \rightarrow \infty}(\frac{ n }{ 1 + n })^n = e\]

Answer 2

@mathmath333 here u go

Answer 3

r u asking its derivation or true or false

Answer 4

as per wolfram the answer is \(\large \dfrac{1}{e}\)

Answer 5

yeah all because
\[\lim_{n \rightarrow \infty }(\frac{ 1+n }{ n })^{-n} = \frac{ 1 }{e } \]

Answer 6

actually its \( \large \lim_{n \rightarrow \infty }\large \left(\dfrac{ 1+n }{ n }\right)^{-n} = e\)

Answer 7

now how is tht possible??

Answer 8

wolfram gave it

Answer 9

as per my experience wolfram is a supercomp and 99.99% correct

Answer 10

nope wolfram is showing 1/e

Answer 11

http://prntscr.com/ak1imr

Answer 12

@mathmath333 we got the values of different questions lol

Answer 13

its the opposite of what u posted

Answer 14

@mathmath333 thanks bruh..I guess I got it ..... though u might get notifications after like 7 hrs after my roommate is awake -_-

Answer 15

i got it now

Answer 16

hold on ....
why do we do tht ??

Answer 17

like to solve tht we convert it in the form
\[\lim_{n \rightarrow \infty}(\frac{ 1+n }{ n })^{-n}~~~then ~~we~~get~~the ~value~~\frac{ 1 }{ e }\]

Answer 18

@mathmath333 u gone -_-

Answer 19

cant we solve tht without converting ??? :/

Answer 20

@mathmath333
nevermind....this time I am sure.... thanks again :))

Answer 21

http://prntscr.com/ak22b2
hav a look ...wht would be the next step ??

Answer 22

it wont even let me Go Pro >_<

Answer 23

do u understand that \(\text{"exp"}\)

Answer 24

yupppp

Answer 25

it means
\[\lim_{n \rightarrow \infty} e^{n \log\frac{ n }{ 1+n }}\]

Answer 26

i forgot that "exp" hence facing difficulty understanding

Answer 27

wht would be the next step ??

Answer 28

@mathmath333 u gotta be kidding me -_- *facepalm

Answer 29

@mathmath333 so it would be
\[\lim_{n \rightarrow \infty} \frac{ e^{n \log n} }{ e^{n \log(n +1)} }\]

Answer 30

lemme know if it makes sense

Answer 31

\(\large \color{black}{\begin{align}
& \lim_{n \rightarrow \infty} \left((\dfrac{ n }{ 1 + n }\right))^{n} \hspace{.33em}\\~\\
& \lim_{n \rightarrow \infty} \text{exp}\log\left(\dfrac{ n }{ 1 + n }\right)^{n} \hspace{.33em}\\~\\
& \lim_{n \rightarrow \infty} \text{exp}n\log\left(\dfrac{ n }{ 1 + n }\right) \hspace{.33em}\\~\\
& \text{exp} \left[\lim_{n \rightarrow \infty} n\log\left(\dfrac{ n }{ 1 + n }\right) \right]\hspace{.33em}\\~\\
\end{align}}\)

try this

Answer 32

@mathmath333 tbh I m blank ..so u need to walk me through this

Answer 33

\(\large \color{black}{\begin{align}
& \lim_{n \rightarrow \infty} \left((\dfrac{ n }{ 1 + n }\right))^{n} \hspace{.33em}\\~\\
& =\lim_{n \rightarrow \infty} \text{exp}\left(\log\left(\dfrac{ n }{ 1 + n }\right)^{n}\right) \hspace{.33em}\\~\\
&= \lim_{n \rightarrow \infty} \text{exp}\left(n\log\left(\dfrac{ n }{ 1 + n }\right)\right) \hspace{.33em}\\~\\
& =\text{exp} \lim_{n \rightarrow \infty} \left(n\log\left(\dfrac{ n }{ 1 + n }\right) \right) \hspace{.33em}\\~\\
\end{align}}\)

Answer 34

okkk go onnn .....I got it till here

Answer 35

@mathmath333 really thanks a lot.... I got it ..... this time its solved for sure .....
again thanks a lot :)) I really appreciate it

Answer 36

bloody latex

Answer 37

lol i was typing, latex betrayed me

Answer 38

just apply L'hopital rest is cake walk

Answer 39

lol yeah .... :P