RSNHS
posted by monique on .
What is the limit of 12x^2 2cosx + cos^2x all over x^2? please answer ASAP. TY

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The limit as x approaches what? Goodness.

There are different ways to approach this:
1. Since both numerator and denominator evaluate to zero as x>0, l'Hôpital's rule applies.
Differentiate the top with respect to x to get:
2cos(x)sin(x)+2sin(x)4x
and the bottom to get
2x
As x>0, both numerator and denominator still > 0, thus we can apply again the rule, and differentiate:
numerator: 2*sin(x)^22*cos(x)^2+2*cos(x)4
denominator: 2
As x>0, the sin(x) term vanishes, and the cos(x) terms cancel out, resulting in 4 over 2 in the denominator.
So the limit is 2.
2. If you have done series expansions before, expand numerator into a power series, taking only terms up to x^4:
12x^22(1x^2/2+x^4/4!)+(1x^2/2+x^4/4!)^2
=(x^824*x^6+144*x^41152*x^2)/576
Dividing by the denominator leaves us with
(x^624*x^4+144*x^21152)/576
and as x>0,
1152/576 = 2 as before.