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\[\cos x \lim_{h \rightarrow 0}\frac{ \sin h }{ h }\]
have you learned lhospital rule ?
no.

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Other answers:

what are you doing in class?
squeeze theorem?
this is one of the last steps in a limit problem from the first unit (limits). it's supposed to evaluate to cosx*1=cosx. I don't understand how sinh/h comes out to be 1?
well, there is a thing called La'Hospital rule, that says when you run a limit and get 0/0 you can take the derivative of the top and the derivative of the bottom and then run the limit again so your limit lim of sin(h)/h = lim of cos(h)/1 = lim cos(h) and at 0 that is 1
l'hopital's rule
lol, we are on the first unit. haven't learned derivates, or any of the theorems or rules.
can someone just explain how the limit of sinh/h is 1?
derivatives*
hmm
do you know the squeeze theorem?
There is a geometric proof http://www.youtube.com/watch?v=Ve99biD1KtA I don't know of an elementary way of showing this limit without geometry and quite a bit of explaining...watch that video.
all right, thanks
np
It appears that you are beginning the study of Calculus with the topic of limits. We often will use a table of values as they approach the limit from the left and from the right. This is the first method you may want to use. A second method that is used is to look at the graph of the function. You can easily see that the limit as h approaches 0 from the left and right is 1. I hope this helps.

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