Final answer:
For (a), the limit is 1/2. For (b), the limit is 0. For (c), the limit is 0.
Step-by-step explanation:
(a) Limit (x → 0) [ (e^x - 1 - x) / x²]:
To find this limit, we can use L'Hôpital's Rule. Differentiating the numerator and denominator with respect to x, we get (e^x - 1 - 1) / (2x). Taking the limit as x approaches 0, we have (1 - 1 - 1) / 0 = 0/0. Applying L'Hôpital's Rule again, we differentiate the numerator and denominator again, giving us e^x / 2. Taking the limit as x approaches 0, we get e^0 / 2 = 1/2.
(b) Limit (x → ∞) x³e^(-x²):
To find this limit, we can use the exponential growth and decay property. As x approaches infinity, the exponential term, e^(-x²), goes to zero. Therefore, the limit becomes (x³)(0) = 0.
(c) Limit (x → 0) [ csc x - cot x ]:
We can simplify this expression using trigonometric identities. csc x is equal to 1/sin x, and cot x is equal to cos x / sin x. Substituting these values, we have 1/sin x - cos x / sin x. Simplifying further, we get (1 - cos x) / sin x. Taking the limit as x approaches 0, we use the fact that sin x / x approaches 1, and cos x approaches 1 as x approaches 0. Therefore, the limit becomes (1 - 1) / 0 = 0/0. To resolve this, we can use L'Hôpital's Rule. Differentiating the numerator and denominator, we get -sin x / cos x. Taking the limit, we have -sin 0 / cos 0 = 0/1 = 0.