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

You got me. Aside from expanding into Taylor series and dividing, I don't come up with a trick to evaluate it.

Taylor series is not taught in our curriculum :-(

The same here :-(

To evaluate the given limit:

Step 1: Start by simplifying the expression. We can simplify the numerator by using the trigonometric identity cos(θ) - 1 = -2sin^2(θ/2). Rewriting the numerator:

lim x->0 [-2sin^2(2x^3/2)] / [sin(2x^6)]

Step 2: Now, let's focus on the denominator. Notice that 2x^6 = (2x^3)^2. To simplify the denominator, we can use the identity sin(θ) / θ = 1. Therefore, we can rewrite the denominator as:

2x^6 / sin(2x^6) = (2x^3)^2 / sin((2x^3)^2)

Step 3: Now, substitute the simplified expressions for the numerator and denominator back into the original expression:

lim x->0 [-2sin^2(2x^3/2)] / [(2x^3)^2 / sin((2x^3)^2)]

Step 4: Apply the limit. As x approaches 0, all the terms of sin(2x^3/2) and sin((2x^3)^2) approach 0 as well. Therefore, we have:

= [-2(0)^2] / [(0)^2 / (0)]

= 0 / 0

Step 5: Since we obtained an indeterminate form of 0/0, we need to apply further techniques to evaluate the limit. In this case, we can use L'Hôpital's rule.

To use L'Hôpital's rule, take the derivative of the numerator and denominator separately with respect to x:

lim x->0 [d/dx (-2sin^2(2x^3/2))] / [d/dx ((2x^3)^2 / sin((2x^3)^2))]

Taking derivatives:

lim x->0 [-4sin(2x^3/2) * (1/2) * cos(2x^3/2) * (6x^2)] / [2(2x^3) / cos((2x^3)^2) * (4x^2 * sin((2x^3)^2))]

Simplifying:

lim x->0 [-12x^2sin(2x^3/2)cos(2x^3/2)] / [4x^2 * 2(2x^3) * sin((2x^3)^2) / cos((2x^3)^2)]

Step 6: Now, substitute x = 0 into the derivative expression:

[-12(0)^2 * sin(2(0)^3/2) * cos(2(0)^3/2)] / [4(0)^2 * 2(2(0)^3) * sin((2(0)^3)^2) / cos((2(0)^3)^2)]

Simplifying:

= 0 / 0

Step 7: We still have the indeterminate form of 0/0 after applying L'Hôpital's rule. We can continue differentiating the numerator and denominator until we no longer have the indeterminate form.

Differentiating again:

lim x->0 [d/dx (-12x^2sin(2x^3/2)cos(2x^3/2))] / [d/dx (4x^2 * 2(2x^3) * sin((2x^3)^2) / cos((2x^3)^2))]

After further differentiation, it is clear that we will still get the indeterminate form of 0/0, so we continue differentiating until the denominator no longer involves a power of x.