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What is the limit of 1-2x^2 -2cosx + cos^2x all over x^2? please answer ASAP. TY

  • Math - trig?? -

    Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.

  • RSNHS -

    The limit as x approaches what? Goodness.

  • Calculus-Limits -

    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)^2-2*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:
    1-2x^2-2(1-x^2/2+x^4/4!)+(1-x^2/2+x^4/4!)^2
    =(x^8-24*x^6+144*x^4-1152*x^2)/576
    Dividing by the denominator leaves us with
    (x^6-24*x^4+144*x^2-1152)/576
    and as x->0,
    -1152/576 = -2 as before.

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