Posted by **monique** on Monday, December 13, 2010 at 6:22am.

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

- Math - trig?? -
**Writeacher**, Monday, December 13, 2010 at 7:19am
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 -
**bobpursley**, Monday, December 13, 2010 at 9:12am
The limit as x approaches what? Goodness.

- Calculus-Limits -
**MathMate**, Monday, December 13, 2010 at 9:18am
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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