>>40896
Ok OP, this stuff is really simple!

Remember, whenever you take the limit of a constant it simply is the constant. also remember that the limits of two things multiplied is equal to the limit of one * the limit of the other, provided they exist (ex: Lim (X*Y) = LIm X * Lim Y)

Because of this, you can just move allt eh constants to the left, which include 1, -1, and 1/ln3. thusly, you move -1/ ln(3) to the left and then simplify the right (x^2/x has one x cancel, since we arent evaluating at 0)

it is (-1/ln3) * (x^2/(x))
since -1/ln3 is constant you can take it outside of the limit
now you have x^2/x inside the limit
since inside the limit x is never 0 you can do the simplify x^2/x to x

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>>40902
>>40903
Thank you so much! while we are at this mind to help me with this anti-derivative?

>>40896
moreover if you were a little shy about canceling an x from the limit, then apply l'Hospital and you'd get (-1/ln(3))* Lim 2x inspected at 0 from the right.

even though it would seem you could then move the 2 over and get a different value (-2/ln3 as opposed to -1/ln3) you'd still notice that lim x is resultant AND the limits for -2/ln3 * lim x at 0+ and -1/ln3 * lim x at 0+ are equal

>>40904
ah, looks like you may have to do a U sub and then a trig sub.

>>40909
>>40904
specifically, i'm seeign that you can set X^2 = u, and then after simplifying you could do 2*sin(t) = u and get a handy form for completing the problem. Just dont forget to switch back!

>>40904
sqrt(4-9x^4) = 2 * sqrt(1 - ((3/2)x^2)^2)

so either do (3/2)x^2 = u then u = sin t
or directly (3/2)x^2 = sin t

3x dx = cos t dt
x dx = (1/3) cos t dt

x dx / 2 * sqrt(1 - ((3/2)x^2)^2)
becomes
(1/3) cos t dt / (2*sqrt(1-(sin t)'2))
since 1 - (sin t)^2 = (cos t)^2
it becomes
(1/3) cos t dt / (2*cos t)
=(1/6) * dt

anti derivative is (1/6) * t = (1/6) * arcsin((3/2)x^2)

that is if I didn't fuck up too much

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>>40997
I did it earlier and i came to the same result.
I have another one i'm stuck with:

>>41013
idea is the same = 2 * sqrt(1-(x/2)^2)
x/2 = sint -> dx = 2*cost dt

2 * sqrt(1-(x/2)^2) dx = 2 * cost * 2*cost dt
= 4 * (cos t)^2 dt (using tri identity)
= 4 * (1 + cos 2t)/2 dt = 2(1 + cos 2t)

(after another substitution u = 2t)
so it is 2t + sin2t = 2t + 2*sint*cost
: 2*arcsin(x/2) + 2*(x/2)*sqrt(1-(x/2)^2)

since sint = x/2 you can compute cost using the identity sin^2+cos^2=1