My Math Forum limit of an intermediate form

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 March 31st, 2010, 05:03 PM #1 Newbie   Joined: Mar 2010 Posts: 1 Thanks: 0 limit of an intermediate form I want to find cos^2 x/2^x as x approaches infinity My first instinct is to use L'Hospitals rule and differentiate unfortunately this leads to -2 sin x cos x/(ln 2 2^x) Which is utterly unhelpful I'm really stuck on this one. If anybody has any insight I would really appreciate BTW from a table of values and a graph I am very confident that the answer is zero
March 31st, 2010, 05:25 PM   #2
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Re: limit of an intermediate form

Quote:
 Originally Posted by atompunk I want to find cos^2 x/2^x as x approaches infinity My first instinct is to use L'Hospitals rule and differentiate unfortunately this leads to -2 sin x cos x/(ln 2 2^x) Which is utterly unhelpful I'm really stuck on this one. If anybody has any insight I would really appreciate BTW from a table of values and a graph I am very confident that the answer is zero
It would be clearer if you put in (). However, it looks to me you have a numerator cos^2(x), which is bounded between 0 and 1 and a denominator 2^x, which becomes infinite with x, so the ratio -> 0.

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