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anonymous
 5 years ago
find the limits when h >0 h /tan^2h (calculus 1 style)
anonymous
 5 years ago
find the limits when h >0 h /tan^2h (calculus 1 style)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think Lokisan did this for you last night. What don't you understand?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because i need to know on the trig way what she told me i understood very clear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i need to do it without the l'hopitals way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0}\frac{h.\cos^2h}{\sin^2h} = \frac{0}{0} = UND\] you have to use L'hopital's rule since you have 0/0 which is an undertermined form.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok,, but the professor told me i have to trig function that can help me to define this sinh /h and cosh1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but why make it so complicated lol? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mm calculus 1... i am not in cal 2... they didnt teach me the l'hopital way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, this is calculus I too :) believe me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0l'hopital is one of the first things taught in Cal I

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you understand L'hopital's rule, then you won't find difficulty solving such problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mm.. so can u teachme de difficult way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i understand it now... but he dont want to look at that in that way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL, I'm afraid my profs skipped the difficult way ^_^" and I'm only aware of this way and many others too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, they skipped the cosh and sinh from our syllabus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, here is the other method (not necessarily recommended for this problem) h lim (h cos h)/h multiplied by lim (h cos h/h) divided by lim (h tan h)/h (htan/h)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why cosh multiplied by cosh

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am just expanding cos^2 h. You can take the lim of one cos h and multiply by another cos h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so, if we change h/tan^2h by h/sin^2h/cos^h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Look this has too many terms to understand this method. This method is usually done for something simple like lim (sin x). If you wanted to us that method on this simple one you multiply by x and divide by x. (You are essentially dividing by 1; it changes nothing.) You are trying to manipulate it because sin x/x =1.
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