find the limits when h ->0 h /tan^2h (calculus 1 style)

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find the limits when h ->0 h /tan^2h (calculus 1 style)

Mathematics
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I think Lokisan did this for you last night. What don't you understand?
because i need to know on the trig way what she told me i understood very clear
but i need to do it without the l'hopitals way

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Other answers:

and i dont know :(
\[\lim_{h \rightarrow 0}\frac{h.\cos^2h}{\sin^2h} = \frac{0}{0} = UND\] you have to use L'hopital's rule since you have 0/0 which is an undertermined form.
ok,, but the professor told me i have to trig function that can help me to define this sinh /h and cosh-1
but why make it so complicated lol? :)
mm calculus 1... i am not in cal 2... they didnt teach me the l'hopital way
:(
lol, this is calculus I too :) believe me
l'hopital is one of the first things taught in Cal I
if you understand L'hopital's rule, then you won't find difficulty solving such problems
mm.. so can u teachme de difficult way
i understand it now... but he dont want to look at that in that way
LOL, I'm afraid my profs skipped the difficult way ^_^" and I'm only aware of this way and many others too
jajaajaja,,,
lol, they skipped the cosh and sinh from our syllabus
:(....
OK, 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)
why cosh multiplied by cosh
I am just expanding cos^2 h. You can take the lim of one cos h and multiply by another cos h
ok so, if we change h/tan^2h by h/sin^2h/cos^h
Look 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.
ohhhhh

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