# L'Hopital's Rule

• Jul 28th 2009, 05:02 PM
ricey
L'Hopital's Rule
lim x-> 0+ (ln cotx)/(e^csc(^2)x) = 0 using L'Hopital's Rule.

Thanks!
• Jul 28th 2009, 05:22 PM
Prove It
Quote:

Originally Posted by ricey
lim x-> 0+ (ln cotx)/(e^csc(^2)x) = 0 using L'Hopital's Rule.

Thanks!

This is tough, but not impossible.

We first need to check that this is of indeterminate form.

It is, since the top $\displaystyle \to \infty$ and the bottom $\displaystyle \to \infty$.

So we need to take the derivative of the top and the derivative of the bottom.

$\displaystyle \frac{d}{dx}[\ln{(\cot{x})}] = -\frac{1}{\sin{x}\cos{x}}$

$\displaystyle \frac{d}{dx}\left[e^{\csc^2{x}}\right] = -\frac{2\cos{x}e^{\csc^2{x}}}{\sin^3{x}}$.

So $\displaystyle \lim_{x \to 0^+}\frac{\ln{(\cot{x})}}{e^{\csc^2{x}}} = \lim_{x \to 0^+}\frac{-\frac{1}{\sin{x}\cos{x}}}{-\frac{2\cos{x}e^{\csc^2{x}}}{\sin^3{x}}}$

$\displaystyle = \lim_{x \to 0^+}\frac{\tan^2{x}}{2e^{\csc^2{x}}}$

As $\displaystyle x \to 0, \csc^2{x} \to \infty$.

So the bottom $\displaystyle \to \infty$.

Therefore this whole limit tends to $\displaystyle \frac{0}{\infty} = 0$.
• Jul 28th 2009, 05:37 PM
ricey
would deriv of ln cot x be sinx/[(cosx )(-sin ^(2)x]

= 1/ [(-cosx) (sinx)]
• Jul 28th 2009, 05:40 PM
Prove It
Quote:

Originally Posted by ricey
would deriv of ln cot x be sinx/[(cosx )(-sin ^(2)x]

= 1/ [(-cosx) (sinx)]

Yes it is.

I'll fix it now...
• Jul 28th 2009, 05:40 PM
ricey
okay phew lol