Question

lim q->1 logq^2(1+qw q-1/q+1)

Answer 1

\[\lim_{q \rightarrow 1} \log _{q^2} (1+qw (\frac{ q-1 }{ q+1 }\]

Answer 2

calculate with l'Hopital for a random \[w \in \mathbb{R} \]

Answer 3

use the property of logarithm first
\(\large \log_mn = \dfrac{\log n}{\log m}\)
what u get ?

Answer 4

\[\frac{ \log(1+qw \frac{ q-1 }{ q+1 }) }{ \log(q^2) }\]

Answer 5

?

Answer 6

yes,
log q^2 = 2 log q , isn't it ?

now you have the form 0/0 and can use L'hopitals,
can you differentiate the numerator and denominator separately ?

Answer 7

log q^2 = 2log q ....?? alright..

Answer 8

yeah, \(\large \log x^m=m\log x\)
thats a standard log property

Answer 9

i get \[\frac{ 1 }{ 1+qw \frac{ q-1 }{ q+1 } } q^2\]

Answer 10

?

Answer 11

derivative of logx= 1/x right?

Answer 12

do you know about chain rule ?

Answer 13

oh right, lemme try this again

Answer 14

so i just take the deriv. of log(..) times the deriv. (..) ?

Answer 15

undefined, is that right?

Answer 16

i get 0 for the num.

Answer 17

hmm ?
as you said, d/dx log x = 1/x

so, in denominator, usingchain rule
(1+qw(q-1/q+1)) times d/dq [(1+qw(q-1/q+1))]

Answer 18

i don't get 0 for numerator...

Answer 19

a little confused, i should review some things

Answer 20

we started with \[\frac{ \log(1+qw \frac{ q-1 }{ q+1 }) }{ 2\log(q) }\]

Answer 21

now i take the derivative of the numerator and denominator separately?

Answer 22

yes,

Answer 23

the numerator derivative will require some algebra

Answer 24

i thought it was 0... i'll try again

Answer 25

taking the derivative of the num. i get \[\frac{ 1 }{ 1+qw(\frac{ q-1 }{ q+1 }) } . (1+qw(\frac{ q-1 }{ q+1 })) \prime\]

Answer 26

am i doing it right?

Answer 27

yup,
w is constant , it will be factored out of deivative.
so, effectively, you have to apply quotient rule for
\(\large w \dfrac{d}{dq}(\dfrac{q^2-q}{q+1})\)

Answer 28

oh wait

Answer 29

applying the quotient rule i get \[\frac{ q+1-q-1 }{ (q+1)^2 }\]

Answer 30

?

Answer 31

i think i'm making a stupid mistake, just don't know what

Answer 32

(2q-1) (q+1) - (q^2-q) = 2q^2 -q-1 -q^2+q = q^2 -1 is what i get for numerator

Answer 33

\[(\frac{ f }{ g }) (a) = \frac{ f'(a)g(a)-f(a)g'(a) }{ (g(a))2 }\]

Answer 34

using this i should get q^2-1?

Answer 35

in the numerator

Answer 36

i'll try again

Answer 37

yes,
derivative of numerator = 2q-1, derivative of numerator = 1

Answer 38

derivative of num. is 2q-1 and 1? huh?

Answer 39

oh, nevermind

Answer 40

yeah, denominator, typo

Answer 41

there's no way i'm getting 2q-1 in the num. ...i'm doing something very wrong with the algebra part

Answer 42

and is it me, or is the algebra part difficult?

Answer 43

num = q^2-q, denom = q+1
derivative of numerator times denominator - derivative of denominator times numerator
= (2q-1) * (q+1) - 1 (q^2-q)

= 2q*q +2q*1 -q*1 -1*1 -q^2 +q

=2q^2-q^2 -q+q +2q -1
= q^2 +2q-1

sorry, i missed the 2q :(

Answer 44

i'm so confused, you're showing how you found the num. using l'hopitals rule, applying quotient rule?

Answer 45

wait i'll write all the steps.

Answer 46

and q^2 +2q-1 is not equal to q^2-2q....

Answer 47

that would help alot

Answer 48

\(\large \frac{ 1 }{ 1+qw(\frac{ q-1 }{ q+1 }) } . (1+qw(\frac{ q-1 }{ q+1 })) \prime \\ \large = \frac{ 1 }{ 1+qw(\frac{ q-1 }{ q+1 }) } .w \dfrac{(2q-1)(q+1)-1(q^2-q)}{(q+1)^2}\\ \large = \frac{ 1 }{ 1+qw(\frac{ q-1 }{ q+1 }) } [w\dfrac{q^2+2q-1}{(q+1)^2}] \)
so thats derivative of numerator,
denominator is easy,
log q^2 =2 log q
its derivative = 2/q in denominator.

so, finally we are here,
\(\large = \frac{ q }{ 2(1+qw(\frac{ q-1 }{ q+1 })) } [w\dfrac{q^2+2q-1}{(q+1)^2}]\)

see whether you have doubts ?

Answer 49

now you can just plug in x=1

Answer 50

***q=1

Answer 51

yup, knew it. i made a stupid mistake

Answer 52

is the answer 1 or w+1?

Answer 53

\(\large = \frac{ q }{ 2(1+qw(\frac{ q-1 }{ q+1 })) } [w\dfrac{q^2+2q-1}{(q+1)^2}]\)
with q=1
\(\large = \frac{ 1 }{ 2 } [w\dfrac{2}{4}]=\dfrac{w}{4}\)
i get w/4

Answer 54

alright, got it. thanks!

Answer 55

welcome ^_^