I just spent almost 40 minutes on a single...

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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?

>>

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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