Question

Help with this L'hospital rule problem:

Solve for the following limit:
Lim as x approaches 0(+) of 2sinxlnx

I don't know what to do, I'm always stuck with a ln0 which is undefined no?

Answer 1

@zepdrix if you don't mind helping me with this one xD

Answer 2

\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\]As x approaches 0 from the right, the limit is approaching,\[\huge 2\cdot 0\cdot -\infty\]
This is of indeterminate form.
But it's not one of the forms we're allowed to apply L'Hop to.
Remember we need one of these two forms,\[\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}\]

Answer 3

We'll have to apply some fancy trig work.\[\large \sin x=\frac{1}{\csc x}\]Substituting gives us,\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\frac{1}{\csc x}}\]

Answer 4

Woops, I wrote that incorrectly :) one sec.

Answer 5

\[\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x\quad=\quad \huge \lim_{x \rightarrow 0^+} \frac{2\cdot \ln x}{\csc x}\]

Answer 6

thats infinity / 0 . we're still not allowed to apply that?

Answer 7

I mean L'hospital rule*

Answer 8

Is it? Hmm lemme think.

Answer 9

oh wait no thats infinity

Answer 10

csc x = 1/sin x, as sin x gets closer to 0, the fraction gets bigger and bigger,

Yah I think you're right, infinity.

Answer 11

That isn't the answer ahaha. I hate cal 1 :(

Answer 12

That's not the right path to take? :o

Answer 13

Seems like we're on the right track :)

Answer 14

I don't know I just have the question, no answer, I dwelled on it for a good hour nothing came to mind. I just know its not infinity :P

Answer 15

* I can verify the answer but it doesn't give it per say

Answer 16

No we're not done yet :) lol

Answer 17

We have successfully gotten it into the form infty / infty, from here we're allowed to apply L'Hop, yes?

Answer 18

yup so I got 2 1/x / csctanx which is 1/sinx * sinx/cos x which gives me 2(1/x)/(1/cosx) am i right?

Answer 19

err i mean -csctanx

Answer 20

-cscx cotx I think

Answer 21

yup -.- so I meant 2(1/x)/-cscxcotx

Answer 22

Which... after you move things around... I think we get,\[\huge \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x}\]

Answer 23

yup and replace sin with 1-cos(square)x sorry french keyboard :P

Answer 24

Hmm it looks like it might still be giving us trouble :( Hmmmm

Answer 25

second derivative?

Answer 26

Its now in a 0/0 form which means that you can apply L'Hop again

Answer 27

But won't it just get worse..? :d

Answer 28

you could use lim x->0 sin(x)/x = 1

Answer 29

Oh derp :o good call.

Answer 30

leaving -2 tan(x) = 0

Answer 31

and phil is right xD it is so hard to see through all that. I wish i could best response both :P

Answer 32

\[\large \lim_{x \rightarrow 0^+}\frac{-2 \cdot \sin^2x}{x \cos x} \quad = \quad -2\lim_{x \rightarrow 0^+}\frac{\sin x}{x} \cdot \lim_{x \rightarrow 0^+} \frac{\sin x}{\cos x}\]

Answer 33

yeah I didn't see that :/

Answer 34

Yah that's neato :O that limit slipped my mind heh

Answer 35

Ah, thats clever