Asked by Aanya

How do we evaluate
lim x-->0 [ cos(2x^3)-1]/[(sin2x^6]

I tried the substitution 2@=2x^3 but that leaves a 'x^2' in the denominator
Answered by Steve
You got me. Aside from expanding into Taylor series and dividing, I don't come up with a trick to evaluate it.
Answered by Anonymo
Taylor series is not taught in our curriculum :-(
Answered by Aanya
The same here :-(
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