Question

Help, problem :
show that lim x to infinity (1+1/x)^e^x = infinity

Answer 1

take the log i guess then take the limit

Answer 2

i.e. take the limit
\[\lim_{x\to \infty}e^x\ln(1+\frac{1}{x})\]

Answer 3

that is in the form \(\infty\times 0\) you can change it to
\[\lim_{x\to\infty}\frac{\ln(1+\frac{1}{x})}{e^{-x}}\] and use l'hopital

Answer 4

What is next after applying l'hopital rule?

Answer 5

what did you get after using l'hospital?

Answer 6

\[\frac{ \frac{ 1 }{ x+1 }- \frac{ 1 }{ x } }{- e ^{-x} }\]

Answer 7

Then i applied it again and got \[\frac{ \frac{ -1 }{ (x+1)^2 } +\frac{ -1 }{ x^2 }}{ e ^{-x} }\]

Answer 8

ok or you could say
\[\frac{\frac{1}{x+1}-\frac{1}{x}}{-e^{-x}}=\frac{\frac{\frac{-1}{x^2}}{1+\frac{1}{x}}}{-e^{-x}}\]

Answer 9

then make that fraction look a lot nicer

Answer 10

\[\frac{\frac{\frac{-1}{x^2} \cdot x^2}{(1+\frac{1}{x})x^2}}{-e^{-x}}\]

Answer 11

\[\frac{\frac{-1}{x^2+x}}{-e^{-x}}\]

Answer 12

now get rid of that compound fraction then you can think about using l'hospital again

Answer 13

just keep using l'hospital rule or when will it actually show that the lim =\[\infty \]

Answer 14

you can keep using l'hosptial if you keep getting 0/0 or inf/inf

Answer 15

you need to clean that one fraction up before applying though

Answer 16

you need to clean so you can see if you can actually use l'hospital

Answer 17

i helped you already clean up your \[\frac{ \frac{ 1 }{ x+1 }- \frac{ 1 }{ x } }{- e ^{-x} } \]
and got \[\frac{\frac{-1}{x^2+x}}{-e^{-x}} \]
all you have to do is get rid of the compound fraction
and see if we have inf/inf or 0/0 then use l'hospital if we do

Answer 18

have you cleaned that fraction up?

Answer 19

working on it :)

Answer 20

i will help just a little more
\[\frac{\frac{-1}{x^2+x}}{-e^{-x}} =\frac{-1}{x^2+x} \div (-e^{-x})=\frac{1}{x^2+x} \cdot \frac{1}{e^{-x}}=?\]

Answer 21

\[\frac{ 1 }{ e ^{-x} (x^2+x)}\]

Answer 22

right?

Answer 23

ok but you could rewrite that as ?

Answer 24

like we plug in really big number we have something that looks like this
1/(0*inf)

that bottom is indeterminate form so you need to rewrite that one fraction above

Answer 25

\[\frac{ 1 }{ xe ^{-x} (x+1)}\] would this make any difference?

Answer 26

nope

Answer 27

1/e^-x=?

Answer 28

ohhhh bring the e^-x up??

Answer 29

e^x/(x^2)+2?

Answer 30

well e^x/(x^2+x)


so what form is this?

Answer 31

\[\infty/\infty \] ?

Answer 32

yep so we can use l'hospital

Answer 33

as you will see it the outcome would look nicer than before

Answer 34

try it

Answer 35

got e^x / 2x+1 , then e^x/2 and it stops there right?

Answer 36

yep
but guess what we have .5*e^x
what do you know about the right end behavior of e^x?

Answer 37

in other words what does e^x approach as x approaches infinity?

Answer 38

where did you get .5*e^x? i graphed e^x and it seems like its approaching infinity

Answer 39

1/2=.5

Answer 40

since e^x approaches infinity as x approaches infinity
then ke^x approaches infinity where k>0 is a constant

Answer 41

That was so difficult, but thank you so much for your help.

Answer 42

your welcome

Answer 43

Glad there are people out there like you willing to spend their time helping out other people!