Question

find the a so the limit is equal to e

lim x -> inf ((x+a)/(x-a))^x
find the a so the limit is equal to e

lim x -> inf ((x+a)/(x-a))^x
@Mathematics

Answer 1

I'm pretty sure that the a should be 1/2 by guessing

but how can I show it?

Answer 2

it is not 1/2

Answer 3

oh...sorry misread the problem...

Answer 4

I was just taking the limit...got \[e^{2a}\]

Answer 5

so if a is 1/2 then the limit would be e like I need right?

Answer 6

I thought that is what you wanted...clearly 1/2 is the answer

Answer 7

I guessed to get that
I don't know how to solve the problem, I only have the solution

Answer 8

I want to know how I can find that solution from the problem

Answer 9

wait how did you evaluate the limit?

Answer 10

are you allowed to use \[\lim_{x\to\infty}\left(1+\frac{r}{x}\right)^x=e^r\]

Answer 11

is that an identity?

Answer 12

yes

Answer 13

I've never seen that before
I think we're supposed to be using l'hopital's rule to solve this

Answer 14

normally written

\[\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x\]

Answer 15

is there a way to solve it without using that? or can you show why that identity works?

Answer 16

have you seen this

\[\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e\]

this is a definition of e

Answer 17

yeah I have seen that before

Answer 18

I can derive the identity from there ,but if you are supposed to use l'hospitals rule we can do that

Answer 19

well that's what we learned this week so could we please do it that way?

Answer 20

I think I've made some progress I have\[\lim_{x \rightarrow \infty} e ^{xln(x+a/x-a)}\]

Answer 21

write

\[\left(\frac{x+a}{x-a}\right)^x\]

\[e^{x\ln\left(\frac{x+a}{x-a}\right)}\]

Answer 22

good

Answer 23

then was thinking that the top needed to be 1 as x->inf

Answer 24

so I focused on that

Answer 25

but the limit of that is indeterminate

Answer 26

but not in the right way to use l'hopital's rule I don't think

Answer 27

\[e^{x\ln\left(\frac{x+a}{x-a}\right)}\]

\[=e^{\frac{\ln\left(\frac{x+a}{x-a}\right)}{1/x}}\]

Answer 28

compute

\[\lim_{x\to\infty}=\frac{\ln\left(\frac{x+a}{x-a}\right)}{1/x}\]

Answer 29

and that puts in 0/0 form right?

Answer 30

if you do it right you will get 2a .... and so the final answer will be \[e^{2a}\]

Answer 31

yes

Answer 32

can you show me how to take the derivative of that?

Answer 33

things don't seem to be getting any simpler
the quotient rule means that I keep the ln(x+a/x-a)

Answer 34

then I just have a big mess from the ln(x+a/x-a) which also uses the quotient rule

Answer 35

write \[\frac{x+a}{x-a}\]

as \[\frac{x+a}{x-a}=\frac{x-a+2a}{x-a}=1+\frac{2a}{x-a}\]

Answer 36

oh good idea I didn't think of doing that

Answer 37

\[\frac{d}{dx}\ln\left(1+\frac{2a}{x-a}\right)\]

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

Answer 38

\[\frac{d}{dx}\frac{1}{x}=\frac{-1}{x^2}\]

Answer 39

put together...

Answer 40

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

Answer 41

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

\[=\frac{1}{1+\frac{2a}{x-a}}\times2a\times\frac{x^2}{(x-a)^2}\]

Answer 42

now take limit

Answer 43

good job in turning that warlock into a princess

Answer 44

lol

Answer 45

really that was amazing

Answer 46

more and more practice you will get better at being manipulative

Answer 47

but for now it looks like magic lol

Answer 48

you understand what I typed?

Answer 49

well turning a warlock into a princess is just a trick its not really magic

Answer 50

I do have one question actually, how come you didn't need to use the quotient rule when you put it together?

Answer 51

you just took the derivative of the top and put it over the derivative of the bottom

Answer 52

That is L'Hospitals rule

Answer 53

ohh yeah, no wonder I was so messed up haha

Answer 54

;)

Answer 55

it all makes sense now! Thank you so much!

Answer 56

good

Answer 57

you can ask zarkon any math question
hes a math god

Answer 58

he really is lol

Answer 59

Demigod

Answer 60

there are still things I don't know

Answer 61

not half
maybe three fourths

Answer 62

lol..I'll take that

Answer 63

\[\frac{3}{4}-god\]

Answer 64

oops that looks like a minus sign

Answer 65

it does

Answer 66

it was suppose to be a dash

Answer 67

you know anything about Diffie-Hellman tuples?

Answer 68

I'm not familiar with Diffie-Hellman (I know what a tuple is ) :)

Answer 69

there is alittle more to it
but i know it has something to do with cyclic groups

Answer 70

ok...I know what a cyclic group is

Answer 71

is this for a class you are taking?

Answer 72

http://en.wikipedia.org/wiki/Computational_Diffie%E2%80%93Hellman_assumption
here is a little more info on it
---
i'm reading a paper

Answer 73

i mean you don't have to look at that site if you aren't interested

Answer 74

interesting

Answer 75

yes it is

Answer 76

zarkon, are you still here? I have another question that I could use some help on

Answer 77

I might be here

Answer 78

what kind of question?

Answer 79

and i know i'm not zarkon lol

Answer 80

tag team

Answer 81

Prove the identity\[\arcsin (x-1/x+1) = 2\arctan \sqrt{x}-\pi/2\]

Answer 82

oh i did that

Answer 83

i can find it

Answer 84

one sec

Answer 85

yep I remember seeing this problem just a little while ago

Answer 86

james did this the cal way
i did it the trig way
look at both and pick what you like
http://openstudy.com/#/updates/4ec58ab6e4b03063799de754

Answer 87

oo thank you

Answer 88

do you have any questions on anything james or i did?

Answer 89

i hope you like the trig way better because i think trig is the freaking bomb!

Answer 90

I like the way you did it
it was how I started doing the problem when I got stuck.

Answer 91

calc way looks cool too though

Answer 92

yeah its okay
its like tapping out at a wrestling match but whatever you like

Answer 93

i was jk

Answer 94

the trig way just seems to work out so much better and easier,
but you can get lost in all those trig formulas if you aren't careful lol

Answer 95

right

Answer 96

i bet my trig students wouldn't be able to figure this one out

Answer 97

the arcsin and arctan make it tricky. I just recently learned that right triangle trick too.

Answer 98

next semester i 'm going give to my trig and cal students as a project
muhahaha however the evil laugh goes