Infinite loop with l'hopital's rule?

[darthxepher]

I'm not supposed to get help with the exact problem, but, in general, you what do if this happens?

 

[Raziel2701]

Are there any powers involved? I think you can take the log of the original function to rewrite the expression into a form where you can use L'Hospitals in a way that works.

 

[Mentallic]

Would it be infinite loops in regards to trigonometry? Give us an indication of the type of problem it is.

 

[darthxepher]

The actual problem is lim x-->0⁺ (cot(x) - 1/x)

Do you think you could come up with a similar problem and help me with it so I can apply that to the original one? Sorry, I think this is making it overly difficult, but I would rather not cheat.

 

[Mentallic]

Mmm... yes that is quite a little toughy. But then again I haven't studied limits passed the high school level so I wouldn't be too sure.

I've tried something, maybe it's what you're looking for? Or even then it could be completely invalid so don't take this solution seriously.

I'll use a similar example of \lim_{x\to 0}\csc(x)-\cot(x)

If we let \lim_{x\to 0}\csc(x)-\cot(x)=k and manipulating the limit gives us \lim_{x\to 0}\frac{\tan(x)-\sin(x)}{\sin(x)\cdot \tan(x)} which is the same as \lim_{x\to 0}\frac{\tan(x)}{\sin(x)}\cdot \frac{1}{\tan(x)}-\frac{\sin(x)}{\tan(x)}\cdot \frac{1}{\sin(x)} and we know that...

edit: felt like I was giving away too much info - since it was the entire solution

Can you finish this off?

 

[sakodo]

Regarding to your problem,

Make the denominator common, apply L'hopital's rule once and after applying a trig identity you should be able to cancel out a tan(x). The answer is pretty clear afterwards.

I will post a solution up if you need.

Regards,

sakodo

 

[Bohrok]

After you combine the fractions in the original problem, you just need l'Hôpital's rule twice to get the limit; no infinite loop