## MarcLeclair 2 years ago 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?

1. MarcLeclair

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

2. zepdrix

$\huge \lim_{x \rightarrow 0^+} 2\sin x \cdot \ln x$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,$\huge \frac{0}{0}, \qquad \frac{\infty}{\infty}$

3. zepdrix

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

4. zepdrix

Woops, I wrote that incorrectly :) one sec.

5. zepdrix

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

6. MarcLeclair

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

7. MarcLeclair

I mean L'hospital rule*

8. zepdrix

Is it? Hmm lemme think.

9. MarcLeclair

oh wait no thats infinity

10. zepdrix

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.

11. MarcLeclair

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

12. zepdrix

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

13. zepdrix

Seems like we're on the right track :)

14. MarcLeclair

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

15. MarcLeclair

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

16. zepdrix

No we're not done yet :) lol

17. zepdrix

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

18. MarcLeclair

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?

19. MarcLeclair

err i mean -csctanx

20. zepdrix

-cscx cotx I think

21. MarcLeclair

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

22. zepdrix

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

23. MarcLeclair

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

24. zepdrix

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

25. MarcLeclair

second derivative?

26. KennethR

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

27. zepdrix

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

28. phi

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

29. zepdrix

Oh derp :o good call.

30. phi

leaving -2 tan(x) = 0

31. MarcLeclair

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

32. zepdrix

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

33. MarcLeclair

yeah I didn't see that :/

34. zepdrix

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

35. KennethR

Ah, thats clever