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