Question

Limit x tends to 0. (2^x - 3^x)*x /(1-cos3x)

Answer 1

for x = 0 we get an indeterminate value 0/0 so we can apply l'hopital' s rule

Answer 2

L'H'S twice seems like something I would try.

Answer 3

the second derivative of 1-cos(3x) is
1-cos(3x)
3sin(3x)
9cos(3x)

the second derivative of 2^x-3^x is
2^x (ln 2)^2 - 3^x (ln 3)^2

so

Lim 2^x (ln 2)^2 - 3^x (ln 3)^2
x -> 0 ---------------------------
9cos(3x)

and then you can plug in 0 for x

Answer 4

no wrong

Answer 5

because
2^x (ln 2) - 3^x (ln 3)
at x=0
is not equal to 0

Answer 6

then we couldn't use LHS twice, sorry

Answer 7

\[\lim_{x \rightarrow 0}\frac{x(2^x-3^x)}{1-\cos(3x)}\]wait, was it this, I made a mistake on interpretation of the problem.

Answer 8

why did you think it necessary to apply LH twice?

Answer 9

oh I see why - the denominator would be 3 sin3x which would give a denominator of 0

Answer 10

Yes fB you right with the problem

Answer 11

I got the following with applying LH twice

-2 * 3^x ln3 - 3^x* x ln^2 (3) + 2^x* x ln^2(2) + 2^(x +1) ln 2
---------------------------------------------------------
9 cos (3x)

Answer 12

- that was a long winded one
I'll check it out on wolfram

Answer 13

Thanks .the answer given in the book is (2/9) log(2/3)

Answer 14

yes that looks good
now plug in x = 0
for your limit

Answer 15

yw

Answer 16

yes

plugging in 0 the 2 middle terms = 0 and denominator = 9

so we have -2 ln3 + 2 ln2
------------
9

== 2ln 2 - 2 ln3
---------
9

= (2/9) ln (2/3)

Answer 17

yw

Answer 18

that was a tricky differentiation

Answer 19

God bless !!! Thanks a ton both you'll guys !!!!!

Answer 20

yw