Question

l'hospital's rule question: I've reduced lim(x->a)(cosx * ln (x-a))/(ln(e^x-e^a) to -infinity/-infinity, can someone verify that?

Answer 1

yes since ln(0) tends to -infinity ....this is now an indeterminate form , so differentiate top and bottom

Answer 2

hmm derivative doesn't look much better though

Answer 3

Any clue what I'll come up with after differentiating everything? I'm getting 0/e^x for a not = to 0.

Answer 4

but I guess, a = 0 would be okay too.

Answer 5

it's zero, isn't it?

Answer 6

does it have to have a value? Can it not be equal to zero? Teacher didn't really teach l'hospital's rule... I don't think I get this teacher, anyway.

Answer 7

well after differentiating, you still get an indeterminate form ..... 0/0

"a" can be anything , its just a constant

Answer 8

most likely your answer will be in terms of "a"

Answer 9

Hmmmm... okay, so I guess I differentiated it poorly, maybe I looked at a as a variable? I've got to go back to this problem and process it again.

Answer 10

Nope, I think I treated it correctly, so I have as f'(x)= ((-sin(x)*ln(x-a)+(cos(x))/(x-a))(e^x-e^a))/e^x

Answer 11

and I've simplified that to 0/e^x

Answer 12

right? cos (x) is -sin(x), ln (x-a) is 1/(x-a)*1, ooh, I don't know how I got ln in the numerator up above.

Answer 13

oh yeah, product rule!

Answer 14

yeah looks ok ... the denominator should be e^x / (e^x - e^a) is that what you have?

Answer 15

Yes, but after I inverted it, I put (e^x-e^a) in my numerator.

Answer 16

so my numerator is: ((-sin(x)*ln(x-a)+(cos(x))/(x-a))(e^x-e^a))

Answer 17

and my denominator is just e^x

Answer 18

right .... ok i don't see how that simplifies to 0/e^x though

Answer 19

when I sub in my limit as e^x goes to a, it also goes to e^a, so it's like e^a-e^a=0. 0 * the rest of the crap in the numerator gives me zero over e^x

Answer 20

the "rest of the crap" :) does not give zero though

--> cos(a) /0 - sin(a) *ln(0)

Answer 21

so what i mean is ...you get 0 on bottom too because of the (x-a)

Answer 22

there's no x -> a on bottom (I moved it to the numerator, that how I've got zero times the rest of the numerator which can only equal zero). Right?

Answer 23

and if e^x goes to e^a, it's still only e^a. The numerator went to zero because it was all getting multiplied by (e^a-e^a). I KNOW I am wrong, can you help me understand why?

Answer 24

in the numerator you have term ... cos(x)/(x-a) its still a fraction so technically the (x-a) is being multiplied by e^x

Answer 25

Okay, so you're saying the numerator is a fraction, I get that, but even so, it's still being multiplied by e^x-e^a. Right? How is it being multiplied by e^x only?

Answer 26

best way to see it is write expression as single fraction
\[[\frac{\cos x}{x-a} -\sin x \ln (x-a)]*\frac{e^{x} -e^{a}}{e^{x}} = \frac{(\cos x - (x-a) \sin x \ln(x-a))*(e^{x} -e^{a})}{(x-a) e^{x}}\]

Answer 27

I see!!! Brilliant!

Answer 28

so 0/0, differentiate s'more?

Answer 29

unfortunately this means we need to differentiate again :{

Answer 30

btw wolfram can't even do this

http://www.wolframalpha.com/input/?i=lim+cos%28x%29*log%28x-a%29+%2F+log%28e^x+-e^a%29+as+x-%3Ea

but i think the answer is ...cos(a)

Answer 31

I'm on it! I'm going to close this thread for now because I have to leave, but I'll be back at it in another hour (and will likely have to post another thread 'cause I'll get stuck again). You're the man/ma'am. Medal for you!

Answer 32

Perfect, thanks!

Answer 33

thanks

Answer 34

Okay, so I JUST can't get the solution that you got. when I combine ...

Answer 35

cos(x)/(x-a)-sin(x)*ln(x-a) and all of that multiplied by (e^x-e^a)/e^x

Answer 36

I instead get: (cos(x)(e^x-e^a)-sin(x)*ln(x-a)(e^x-e^a)(x-a))/e^x(x-a)

Answer 37

now that simplifies very easily to 0/0, but how on Earth am I going to differentiate this mess?

Answer 38

oh wow, yeah it gets messy ...but just use the product rule a bunch of times

notice the denominator will become e^x + e^x(x-a) after differntiation

Answer 39

so that means no "0" on bottom so thats good

Answer 40

haha i worked out the derivative before but i've erased it .... somehow it simplifies to cos(a)

Answer 41

also i noticed the only difference in my fraction and yours is i factored out the (e^x -e^a) in numerator