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as x approaches 0?

How can a limit not exist? limit of ((x^2)^1/2) / (x + 2x^2) as x approaches 0 = x / ( x (1+2x) ) =0 or dne

Thanks a lot.

lim = -1/2 as x approaches 0

lim = 0

-sin^2 / (sin^2 + X^3(cos +1)) = 1/(1+ (x^3(cos +1))/sin^2) lim= 1

1+ (sin^2/x^3(cosx+1)) limit of cos + 1 = 2

(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3) limit =1

- 1/cos + 1 + sin^2/x^3 cos x + sin^2/X^3 limit = 1?

i still get 0/0

cos^2 - 1 / ( sin^2 + x^3)(cos x + 1) = - sin^2/ (sin^2 + x^3 )(cos +1) is this the right way?

I know l hospitals rule u take the derivative and then find the limit but is there another way u can so it by using sinx/ x = 1

Homework Statement lim (cos x - 1) / (sin^2 x + x^3) as x approaches 0. Homework Equations sinx/x = 1 The Attempt at a Solution I get 0/0. Is that the answer?