Question

\[\lim_{x\rightarrow \infty}{(1+\frac{3}{x})^{4x}}

Answer 1

\[\lim_{x\rightarrow \infty}{(1+\frac{3}{x})^{4x}}\]

Answer 2

\[\lim_{x \to \infty} \left( 1 + \frac{3}{x}\right)^{4x}\]

Answer 3

hello

Answer 4

we know that ( 1 + 1/x ) ^x as x->oo is e

Answer 5

I didn't know that actually :-P

Answer 6

so me and the rest of those who know ;)

Answer 7

just pretend that you know ;)

Answer 8

what rules are you supposed to use anyway ?

Answer 9

I have no idea... it's a question that tests the convergence of a sequence.... and I am sure it does but I don't know the value it converges to.

Answer 10

lim ( 1 + 3/x) ^(4x) = [ 1 + 1 / ( x/3) ] ^4x

Answer 11

let u = x/3, as x->00, u->oo because oo/3 ->oo

Answer 12

true

Answer 13

lim ( 1 + 1/u ) ^ ( 4*3u)

Answer 14

lim [( 1 +1/u)^u ]^12 , so it will be e^12

Answer 15

that's the answer

Answer 16

Hey Perl
If f(x) -> o
(1 + f(x))^(g(x)) where g(x) tends to infinity
it is equal to e^(g(x)*ln f(x)) right?

Answer 17

I still don't get it :-(

Answer 18

ishann, it depends

Answer 19

but youre right insofar as its an indeterminate form

Answer 20

agdg, basically 'we know' that ( 1 +1/x) ^x goes to e as x->oo, and we just use a substitution

Answer 21

agdg, it is usually given by the teacher

Answer 22

given : lim ( 1 +1/x) ^x as x->00 is e, this should be in your notes.
you could try saying assume it has a limit , so y = ( 1 + 1/x) ^x , then
ln y = x ln ( 1 + 1/x)

Answer 23

find the limit ln ( 1 + 1/x) / ( 1/x) as x->oo

Answer 24

you have 0/0, so now use Lhopital's rule

Answer 25

oooh alright

Answer 26

1/ ( 1 + x^-1) *-1*x^-2 * 1/ (-1*x^-2) now those cancel so we have

Answer 27

you are left with 1 / ( 1 + 1/x) , and the limit of that as x->oo is 1 , so ln y = 1, so y = e^1

Answer 28

now that we know lim (1 + 1/x) ^x is e, we can use a substitution to solve the original problem

Answer 29

yay I will impress my calc teacher thanks now I get it!

Answer 30

ishann, can you link your pdf

Answer 31

thats a powerful theorem, im not sure though it works for all cases

Answer 32

If f(x) -> o
(1 + f(x))^(g(x)) where g(x) tends to infinity
it is equal to e^(g(x)*ln f(x))...
only under certain conditions

Answer 33

also im not sure how that helps in this problem

Answer 34

e^( 4x * ln ( 1/x) ) ?

Answer 35

e^(f(x)*g(x)) that is what I mean not the ln part sorry

Answer 36

take the log, take the limit, exponentiate although there is probably a snappier way to do it with your eyes

Answer 37

no you were right before,
we were given f(x)^g(x), so we know that f(x)^g(x) = e^[ ln f(x)^g(x)]

Answer 38

so f(x)^g(x) = e^[ g(x) ln f(x) ]

Answer 39

if you want a guess i would say it is
\[e^{12}\] it is surely e to the something

Answer 40

so find lim e^[ 4x ln ( 1 + 1/x) ]

Answer 41

so that doesnt really get us anywhere

Answer 42

not it can't be like that in my opinion

y = (1 + f(x))^g(x)
f(x) tends to zero while g(x) to infinity
logy = g(x)log(1 + f(x))
but since f(x) tends to zero it must become
log(1 + f(x)) tends to f(x)
log y = g(x)*f(x)
y = e^(f(x)*g(x))

Answer 43

start with
\[4x\ln(1+\frac{x}{3})\] then rewrite as
\[4\frac{\ln(\frac{3+x}{3})}{\frac{1}{x}}\] and use l\hopital

Answer 44

ishaan, but then you would get e^4 , which is false

Answer 45

log 1 = 0
but there we have 1 + f(x) where f(x) tends to zero it is near to zero but not zero so it has to be f(x) for the whole log(1 + f(x))

Answer 46

e^( 1/x * 4x ) = e^4

Answer 47

\[e^{\frac{3}{x}*4x} = e^{12}\]

Answer 48

but since f(x) tends to zero it must become
log(1 + f(x)) tends to f(x) ?

Answer 49

log ( 1 + f(x)) ~ f(x) if f(x)->0 ?

Answer 50

yeah that is what I am thinking

Answer 51

it must be like that

Answer 52

i disagree

Answer 53

lim log ( 1 + u) / u as u->0 is not 1

Answer 54

err, wait it is

Answer 55

lim log ( 1 + x ) / x as x->0 is 1, so they are asymptotic to each other. very good, you used a different fact

Answer 56

but to prove this we use Lhopital, lim 1/(1+x) / (1) and as u->0 it becomes 1

Answer 57

right! nice

Answer 58

i took derivative of top and bottom

Answer 59

interesting argument , you just thought of that on the fly ?

Answer 60

oh and f(x) must be continuous , etc

Answer 61

I saw that somewhere couldn't recall it, but yeah the proof just bumped in my head

Answer 62

hey how'd you track katrina's ip?

Answer 63

hmmm, shivampatel.net

Answer 64

dont tell her though

Answer 65

but, when i compare log ( 1 + x) to x , i get a discrepanct

Answer 66

discrepancy *

Answer 67

Okay! I won't; Thanks

Answer 68

hmmm, maybe it should be ln ( 1 + x) , not base 10

Answer 69

ok perfect, now it works

Answer 70

yes now it is VERY close , it has to be ln , not log (base 10 )

Answer 71

yeah natural log

Answer 72

so ln ( 1 + u ) ~ u for u -> 0 , and you can substitute a continuous function there

Answer 73

you did a very nice argument

Answer 74

but its false that log ( 1 + u ) ~ u , for base 10

Answer 75

change it to log ( 1 + u ) ~ ln ( 10 ) u ?

Answer 76

oh i see my mistake , the derivative of log ( 1 + x ) is not 1/(1+x) if it is not base e

Answer 77

thanks you made it better

Answer 78

log ( 1+u) = ln ( 1 + u ) / ln 10 ~ u / ln 10

Answer 79

there, fixed it now

Answer 80

log ( 1 + u) ~ u/ln 10 as u-> 0+

Answer 81

excellent perl
I presume you are university student

Answer 82

i have bachelor's ;)

Answer 83

wow awesome in math right?

Answer 84

*shrugs*

Answer 85

i love math, it keeps me alive ;)

Answer 86

i suffer from depression , it is my joy

Answer 87

i love math too I wanna do game programming

Answer 88

nice loving math is a good thing to do

Answer 89

i like world at war, very nice maps

Answer 90

call of duty

Answer 91

some maps look foggy and crappy, but some are pretty

Answer 92

yes Indeed game programming uses lot of math if you like games you can get started too

Answer 93

oh, hey do you like 3d graphing? i found a program
romanlabs.com and a crack

Answer 94

I mean you will have upper hand you are good in maths all you need to learn is just the language

3d graphing yeah though I didn't do much of 3d math but yeah i like it thanks for the resource

Answer 95

hey is this a joke
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Answer 96

10 = 2 in binary math

Answer 97

lol

Answer 98

looks like 3 to me