answer:To evaluate this limit, we can use L'Hopital's rule. L'Hopital's rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator. In this case, we can see that the limit of the numerator and denominator is both 0 as x approaches 1. Therefore, we can use L'Hopital's rule to evaluate the limit. The derivative of the numerator is -2x and the derivative of the denominator is πcos(πx). Therefore, we have: lim_{xto1}dfrac{1-x^2}{sin (π x)} = lim_{x to 1}frac{-2x}{pi cos (pi x)} = frac{-2}{-pi} = frac{2}{pi} Therefore, the limit of the function as x approaches 1 is 2/π.

answer:Since the sine function is periodic with period (2pi), we can add or subtract multiples of (2pi) to the argument of the sine function without changing its value. Therefore, we can rewrite the expression as sin^{-1}left[sinleft(-frac{7pi}{6}right)right] = sin^{-1}left[sinleft(-frac{7pi}{6} + 2piright)right] = sin^{-1}left[sinleft(frac{pi}{6}right)right] = sin^{-1}left(frac{1}{2}right) = frac{pi}{6} Therefore, the exact value of the expression is (frac{pi}{6}).

answer:Yes, the inequality is true in this case. This is because the Cauchy-Schwarz inequality implies that |x_ioverline{y_i}| leq ||x_i|| ||y_i||, and since the eigenvectors are orthogonal, we have ||u_i||^2 = ||x_i||^2 + ||y_i||^2. Substituting these inequalities into the original expression gives the desired result.

answer:The sum can be simplified as follows: [ sumlimits_{i=0}^{|k|} i^2 + 2i = sumlimits_{i=0}^{|k|} (i^2 + 2i) = sumlimits_{i=0}^{|k|} i(i + 2) ] Let's consider a specific case with i = 1: [ 1(1 + 2) = 3 ] Since 3 is a prime number, we have found an integer k = 1 for which the sum is prime. The general term for the sum is given by: [ frac{k(k + 1)(2k + 7)}{6} ] For k > 6, the sum will always be composite, as it can be divided by 6 (since the factors will include at least one multiple of 3 and another multiple of 2). Therefore, the sum is prime only for k = 1 and k = 2. The original confusion was about the interpretation of the sum and the conditions for existence. It's important to note that only one counterexample is needed to establish the existence of a prime number under the given conditions.