Find the value of \( \lim_{x \to 0} \left( \frac{x^2+2\cos x-2}{x\sin^3 x} \right) \)
Step-by-step Solution:
We can evaluate this limit using the Taylor series expansion for \( \cos x \).\n\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \]\n\[ = 1 - \frac{x^2}{2} + \frac{x^4}{24} - O(x^6) \]\n\nSubstitute this into the numerator:\n\[ \text{Numerator} = x^2 + 2\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\right) - 2 \]\n\[ = x^2 + 2 - x^2 + \frac{x^4}{12} - 2 \]\n\[ = \frac{x^4}{12} + O(x^6) \]\n\nFor the denominator, use the standard limit property \( \sin x \approx x \) as \( x \to 0 \):\n\[ \text{Denominator} = x\sin^3 x \approx x(x)^3 = x^4 \]\n\nNow compute the limit:\n\[ \lim_{x \to 0} \frac{\frac{x^4}{12} + O(x^6)}{x^4} = \frac{1}{12} \]