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I just spent almost 40 minutes on a single calculus exam problem. It was the last question too.
lim x->0 ( (3x^2 - 1) * cos^2(x) + 1 ) / ( 3x^2) )
I was able to plug in that cos^2(x) = 1 - sin^2(x). After that I had no fucking clue.
Calc-bros of /sci/, tell me how much of an idiot I am?
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Sin(x)/x
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>>7843851
which is 1 right?
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>>7843854
Yea
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>>7843854
Yes, which is not hard to prove with squeeze theorem
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>mfw I don't understand any of this shit as I'm still learning quadratics
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>>7843850
2/3 right ?
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>>7843850
The choices were
a. 4/3
b. infinity
c. does not exist
d. -1
e. none
I ended up choosing d.
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>>7843875
4/3 is right
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Distribute the (cos(x))^2
cancel the two 3x^2
which = 1 and it become lim x->0 1 = ?
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>>7843877
how did you get that? I must have tried half a dozen methods of attack, and nothing seemed right.
Does it involve taking a conjugate?
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>>7843850
[eqn] 3x^2 \cos ^2 (x) - \cos ^2 (x) + 1 = 3x^2 (1- \sin ^2 (x) ) - 1 +sin ^2(x) +1 = 3x^2 (1- \sin ^2 (x) ) +sin ^2(x) = 3x^2 - 3x^2 \sin ^2 (x) + \sin ^2 (x) = 3x^2 (1- 2 \sin ^2 (x) ) [/eqn] Cancel numerator and denominator. Limit is 1
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>>7843900
no it's 4/3
http://www.wolframalpha.com/input/?i=lim+x-%3E0+(+((3x%5E2+-+1)+*+cos%5E2(x)+%2B+1+)+%2F+(+3x%5E2)+)
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>>7843899
seems so obvious now.
thanks
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>>7843921
I used your hunch so I guess it was a good start
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