Question

Determine the values of "a" and "b" for which the following limit is true:

limit as x approaches infinity of
((ax+1)/(bx-1))^x = 4

Answer 1

\[\left(\frac{ax+1}{bx-1}\right)^x=4\]?

Answer 2

yes

Answer 3

find the limit in \(a\) and \(b\)

Answer 4

how do you do that?

Answer 5

oh this is a pain
start with by taking the log to get \[x\ln\left(\frac{ax+1}{bx-1}\right)\]

Answer 6

ok, and the other side would equal ln(4), right?

Answer 7

wait, no it wouldn't

Answer 8

yeah now i am confused

Answer 9

we need to compute that limit in \(a\) and \(b\) and it is not of the form to use l'hopital because we don't know if \(a>b\) or if \(a<b\) or what

Answer 10

so now i am thinking this method may not be right

Answer 11

why would you have to know whether a >b or not to use l'hopital's rule?

Answer 12

wouldn't l'hopital's rule work for any values of a and b?

Answer 13

can anyone else help? satellite73 went offline....

Answer 14

I dont know any of this the most i can do is wolfram alpha

Answer 15

\(a=b=-\ln 4\) seems to work

Answer 16

how did you get that?

Answer 17

that doesn't work, try this :
\(a=b = \dfrac{1}{\ln 4}\)

Answer 18

1/ln4 gives me 16
http://www.wolframalpha.com/input/?i=lim%28x%5Cto%5Cinfty%29+%28%281%2Fln%284%29*x%2B1%29%2F%281%2Fln%284%29*x-1%29%29%5Ex

Answer 19

so that doesn't work either.. wait il explain once i figure it out

Answer 20

ok

Answer 21

it is \(a=b=\dfrac{1}{\ln 2}\) !

http://www.wolframalpha.com/input/?i=lim%28x%5Cto%5Cinfty%29+%28%281%2Fln%282%29*x%2B1%29%2F%281%2Fln%282%29*x-1%29%29%5Ex

Answer 22

ok, is there any way using calculus to get that answer?

Answer 23

remember how to do long division ?

Answer 24

yes

Answer 25

\[\dfrac{ax+1}{bx-1}=?\]

Answer 26

would that be \[\frac{ a }{ b } + \frac{ 1-\frac{ a }{ b } }{ bx-1 }\] ?

Answer 27

try again

Answer 28

quotient is correct
but remainder doesn't look correct..

Answer 29

bx - 1 | ax + 1 | a/b
ax - a/b
------------------
1 + a/b



(ax+1)/(bx-1) = a/b + (1 + a/b)/(bx-1)

yes ?

Answer 30

true, I messed up while subtracting

Answer 31

\[
\left(\frac{ax+1}{bx-1}\right)^x\\~\\
=\left(\frac{a}{b}+\frac{1+a/b}{bx-1}\right)^x\\~\\


\]

Answer 32

before proceeding further, let me ask you a question

Answer 33

as you make \(n\) large, what number does the below expression approach ?
\[(0.9)^n\]

Answer 34

zero

Answer 35

good. how about below expression :

\[(1.1)^n\]

what number does that approach as you make \(n\) large ?

Answer 36

I presume that should go to infinity, right?

Answer 37

Excellent! as you can see, \((1\pm x)^n\) is either \(0\) or \(\infty\) for nonzero \(x\) as \(n\to\infty\)

Answer 38

Keeping that in mind, lets look at our problem

Answer 39

\[
\left(\frac{ax+1}{bx-1}\right)^x\\~\\
=\left(\frac{a}{b}+\frac{1+a/b}{bx-1}\right)^x\\~\\


\]

Answer 40

As \(x\to \infty\), is there any hope for above expression to be \(4\) if \(\dfrac{a}{b}\) is greater than \(1\) ?

Answer 41

no, because you would be adding a number to 4, so if a/b is greater than 1, than the result will always be greater than 4

Answer 42

In other words, for \(\dfrac{a}{b}\gt 1\), the limit "diverges"

Answer 43

what happens if \(\dfrac{a}{b}\lt 1\) ?

Answer 44

I'm guessing that it would approach 4?

Answer 45

My guess is that it would approach 0

Answer 46

as you can see, i want to argue that \(\dfrac{a}{b}\) has to equal to \(1\)

Answer 47

yes, I understand, if you make b arbitrarily big, then the limit of the whole thing would approach zero, that makes sense

Answer 48

so if the limit does not go to 4 when a/b is greater than 1 or less than 1, that means a/b must equal 1. Which means that a must equal b.

Answer 49

Yes, plug that in \(a=b\)

Answer 50

\[
\left(\frac{ax+1}{bx-1}\right)^x\\~\\
=\left(\frac{a}{b}+\frac{1+a/b}{bx-1}\right)^x\\~\\
\]

\(a=b\), so :
\[
\left(1+\frac{1+1}{bx-1}\right)^x
\]

Answer 51

ok, so that would simplify to \[(1+\frac{ 2 }{ bx-1 })^x\]

Answer 52

would I have to take a common denominator next?

Answer 53

actually, taking a common denominator will bring me back to the original equation

Answer 54

we may use this :
\[\lim\limits_{x\to\infty} (1+\dfrac{t}{x})^x = e^t\]

Answer 55

but it seems we need to massage a bit before using that

Answer 56

ok, and how exactly would we denominator to equal to exponent?

Answer 57

\[
\left(\frac{ax+1}{bx-1}\right)^x\\~\\
=\left(\frac{a}{b}+\frac{1+a/b}{bx-1}\right)^x\\~\\
\]

\(a=b\), so :
\[
=\left(1+\frac{1+1}{bx-1}\right)^x\\~\\
=\left(1+\frac{2/b}{x-1/b}\right)^x\\~\\
\]

Answer 58

fine with above ?

Answer 59

yeah, you just divided both top and bottom with b

Answer 60

way I went about it: \[\lim_{x \rightarrow \infty}(\frac{ax+1}{bx-1})^x=4 \\ \text{ Take }\ln( ) \text{ of both sides: } \\ \ln[\lim_{x \rightarrow \infty}(\frac{ax+1}{bx-1})^x]=\ln(4 ) \\ \text{ you can bring the limit outside } \log \\ \lim_{x \rightarrow \infty} \ln[(\frac{ax+1}{bx-1})^x]=\ln(4) \\ \text{ use power rule for } \log \\ \lim_{x \rightarrow \infty} x \ln (\frac{ax+1}{bx-1})=\ln(4) \\ \lim_{x \rightarrow \infty} \frac{\ln(\frac{ax+1}{bx-1})}{\frac{1}{x}} =\ln(4) \\ \text{ In order for the limit \to actually exist we need } \frac{ax+1}{bx-1} \rightarrow 1 \text{ as } x \rightarrow \infty \\ \text{ so we can have } \frac{0}{0} \\ \text{ and use l'hospital rule } \\ \text{ so this does give you } \frac{a}{b}=1 \implies a=b \\ \text{ So you have } \\ \lim_{x \rightarrow \infty}\frac{\ln(\frac{ax+1}{ax-1})}{\frac{1}{x}}=\ln(4)\]
see if you can finish
apply l'hospital

Answer 61

\[
\left(\frac{ax+1}{bx-1}\right)^x\\~\\
=\left(\frac{a}{b}+\frac{1+a/b}{bx-1}\right)^x\\~\\
\]

\(a=b\), so :
\[
=\left(1+\frac{1+1}{bx-1}\right)^x\\~\\
=\left(1+\frac{2/b}{x-1/b}\right)^x\\~\\
=\left(1+\frac{2/b}{x-1/b}\right)^{x-1/b}*\left(1+\frac{2/b}{x-1/b}\right)^{1/b} \\~\\
\]

taking limit gives you
\[
= e^{2/b}*1
\]

Answer 62

Thank you freckles, I'll try that as soon as I'm done with ganeshie8's way

Answer 63

Ok, so I'm a bit confused as to how the second term equals 1

Answer 64

would it be because as x is approaching infinity (2/b)/(x-1/b) is approaching zero, so 1^(1/b) would just be 1?

Answer 65

Yes. notice that 1/b is a constant with respect to x :

\[\lim\limits_{x\to\infty}\left(1+\frac{2/b}{x-1/b}\right)^{1/b} = \left(\lim\limits_{x\to\infty}1+\frac{2/b}{x-1/b}\right)^{1/b}\\~\\~\\
= (1+0)^{1/b}\\~\\
=1
\]

Answer 66

ok, that makes sense

Answer 67

recall, for continuous functions, we have :
\[\lim f(g(x)) = f(\lim g(x))\]

Answer 68

\[\lim [f(x)]^n = [\lim f(x)]^n\]

Answer 69

ok so lim[e^(2/b)] equals [lim e]^(2/b) ?

Answer 70

No, we're done.

the limit evaluates to \(e^{2/b}\), simply set this equal to \(4\) and solve \(b\)

Answer 71

i got b = 2/ ln(4)

Answer 72

Looks good !

Answer 73

so how does that simplify to 1/ln(2) ?

Answer 74

use below log property :
\[\ln(x^a) = a*\ln(x)\]

Answer 75

ok, so it looks like it should be 1/ln(4^(1/2)), but I'm not sure whether that would be the correct way to do it

Answer 76

Easy...

ln(4) = ln(2^2) = 2ln(2)

Answer 77

because 4 = 2^2

Answer 78

OHHHH, lol, that was literaly the stupidest question I have asked.....

Answer 79

Thank you so much for your help and your time!

Answer 80

haha thats okay :) my stupidiest question was way stupidier than ur stupidiest q

Answer 81

do scroll up and see freckles method using Lhopital rule...

Answer 82

yes, I'm actually looking at his method right now