# someone help me with this calculus problem.

Discussion in 'The Lounge' started by colbystephens, Oct 3, 2006.

1. ### colbystephens1 ton statusGMOTM Winner

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ok, so i'm not really knowing how to go about this problem. we're studying L'Hospital's Rule, so i know that's part of it. i guess i'm just uncertain how to handle the secant. anyway, here's the problem...

lim (sec7x)(cos3x)
as x tends to the negative side of pi/2.

i know the limit is equal to positive infinity, but i'm not sure how to determine that from the equation. i suppose what it comes down to is how do you derive it? both parts require the chain rule, right? if so, does that mean that sec7x goes to 7(sec7xtan7x), and cos goes to 3(-sin3x)?

any help would be appreciated.

2. ### 1979jimmy3501/2 ton status

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could you write sec7x as 1/cos7x giving you co3x/cos7x then you can take the deratives and use LH

3. ### thatK30guy1 ton statusPremium Member

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Greg72 would be the man to help you out with this. Shoot him a PM and post his reply here for us dummies to see.

4. ### Avery4jc1 ton statusPremium MemberGMOTM Winner

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Is Greg a teacher or just oober smart?

-Avery

5. ### thatK30guy1 ton statusPremium Member

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A damn oober smart mofo. He hangs around mainly in the COG and the 67-72 forums.

6. ### ct85k101/2 ton status

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thats what im thinking to but i could be wrong calc 1 was 2 semesters ago

7. ### 1979jimmy3501/2 ton status

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im almost postive thats what should be done, LH is calc 2 which i had in the spring so it is somewhat fresh, im in calc 3 right now

8. ### ARAMP1Aviator Extraordinaire

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The answer is "3/7". 1979Jimmy350 is right on the money.

9. ### ARAMP1Aviator Extraordinaire

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Did you ever get this figured out?

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