1. ## Help with limit

Can anyone show how to solve this?

$\displaystyle \displaystyle\lim_{x\to\{0^+}\frac{ln x}{csc x}$

(Sorry for that weird bracket in the limit... it's just supposed to be "the limit as x goes to 0 from the right.")

Pretty sure I'm supposed to use L'Hopital's rule for this, but when I do, I get stuck after the first derivative. Any advice?

2. It will be (After using LHospitals Rule): $\displaystyle \frac{-1}{xcsc(x)\cot(x)}$

3. When I use L'Hopital's rule I end up with

$\displaystyle \frac{1/x}{-cot(x)csc(x)}$

4. Originally Posted by paupsers
When I use L'Hopital's rule I end up with

$\displaystyle \frac{1/x}{-cot(x)csc(x)}$
Sorry I edited it.

5. Originally Posted by paupsers
When I use L'Hopital's rule I end up with

$\displaystyle \frac{1/x}{-cot(x)csc(x)}$
That is correct, take the derivative again.

probably easier to think of 1/x as x^-1

6. Apply LHospitals Rule again

7. Using the rule again, I end up with

$\displaystyle \frac{-x^-2}{csc(x)(csc^2(x)-cot(x))}$

Plugging in 0 gives me $\displaystyle \frac{0}{\infty(\infty-\infty)}$

Does that equate to 0?

8. infnity - infinity is indeterminated

9. So do I have to use the rule again?

10. Don't make it too complicated:

$\displaystyle \sin (x)\ln (x)=\frac{\sin x}{x}\cdot x\ln x,$ for $\displaystyle x\ne0,$ now take limits, what do you get?

11. Originally Posted by paupsers
So do I have to use the rule again?
Check your derivative for the denominator.
Are you sure your signs are correct?