# Limits and inverse trig functions

• Nov 8th 2012, 11:08 AM
tonykart44
Limits and inverse trig functions
Hi everyone,

I'm currently studying for my calculus test tommorow and I have some problems which I have no clue how to solve. One problem is a limit and the other one is about inverse trig functions.

1) Lim (cos 2x)^(csc^2 2x)

x-> 0

2) 2. Cos(2arccotx)

Any help at all is very much appreciated, as I truly have no idea how to get started on especially the first exercise. Oh and I should mention that we are not allowed to use L'Hopitals rule for these problems.
• Nov 8th 2012, 11:27 AM
Siron
Re: Limits and inverse trig functions
1) $\lim_{x \to 0} \frac{\cos 2x}{\csc^2(2x)}$
Note that the limit is equal to
$\lim_{x \to 0} \cos(2x) \sin^2 (2x)$
2) $2 \cos[2\mbox{arc}\cot(x)]$
Hint: use the double-angle identity for $\cos(2u)$
• Nov 8th 2012, 11:38 AM
tonykart44
Re: Limits and inverse trig functions
Thank you very much for your fast answer! But I think I wasn't clear on the first problem, It should be cos(2x) to the power csc^2 2X Sorry if that wasnt clear.

For the second one I did what you said and I got (2X^2/(1+X^2)) -1 does that look alright to you?
• Nov 8th 2012, 12:06 PM
Siron
Re: Limits and inverse trig functions
No, it's my fault, for some reason I red a fraction in stead of a power.
In case we have to compute $\lim_{x \to 0} \cos(2x)^{\csc^2(2x)}$. Use the fact that $e^{\ln(x)}=x$ thus we can write the limit as $\lim_{x \to 0} e^{\csc^2(x) \ln[\cos(2x)]}$ (note $\ln(a^b) = b \ln(a)$). Because the exponential function is continuous we can write the limit as $e^{\displaystyle \lim_{x \to 0} \frac{\ln[\cos(2x)]}{\sin^2(2x)}$ which can be solved by using l'Hopital's rule.

The solution of the second question is correct. (I assume the 2. was not a factor 2 but just an indication of question 2)
• Nov 8th 2012, 12:23 PM
tonykart44
Re: Limits and inverse trig functions
Once again, I will need to annoy you because for these problems, we are actually not allowed to use L'Hôpitals rule. I'm trying to use trig functions to get around that, but I just can't seem to get that right.

Ok so I got this one in the form (1-2sin^2 x)^(1/2sin^2 2x) but i don't know how to go on from there. I know i'm almost there but i don't know how to get rid of the 2x above, I know it's silly, but i just don't see it.
• Nov 10th 2012, 01:22 AM
Siron
Re: Limits and inverse trig functions
I have no idea how to solve this without l'Hopital's rule.