answer:The problem states to show that the sequence ( frac{a_n}{sqrt{n}} ) is convergent and to find its limit, where the sequence ( a_n ) is defined as: [ a_1 = 1, quad a_2 = sqrt{2 + sqrt{2}}, quad ldots, quad a_n = sqrt{n + sqrt{n + ldots + sqrt{n}}}, ldots ] Note that in ( a_n ), the number of nested square roots is ( n ). 1. **Initial Inequality Insight**: We start by observing that [ a_n = sqrt{n + sqrt{n + ldots + sqrt{n}}} geq sqrt{n}, ] since ( sqrt{x} ) is a strictly increasing function for ( x > 0 ). Thus, [ frac{a_n}{sqrt{n}} geq 1. ] 2. **Upper Bound Establishment**: To establish an upper bound, we show that ( a_n < sqrt{n} + 1 ) for ( n geq 1 ). We will prove this through mathematical induction. Let ( a_k(n) ) denote the expression ( sqrt{n + ldots + sqrt{n}} ) with ( k ) square root signs. We aim to show that ( a_k(n) < sqrt{n} + 1 ) for all ( k ). **Base Case**: For ( k = 1 ), [ a_1(n) = sqrt{n} < sqrt{n} + 1. ] **Inductive Step**: Assume ( a_{k-1}(n) < sqrt{n} + 1 ) holds for some ( k - 1 ). Then, [ a_k(n) = sqrt{n + a_{k-1}(n)} < sqrt{n + sqrt{n} + 1}. ] Next, we show that ( sqrt{n + sqrt{n} + 1} < sqrt{n} + 1 ). Squaring both sides, [ n + sqrt{n} + 1 < (sqrt{n} + 1)^2. ] Simplifying the right-hand side, [ (sqrt{n} + 1)^2 = n + 2sqrt{n} + 1. ] Hence, [ n + sqrt{n} + 1 < n + 2sqrt{n} + 1. ] Since this is clearly true, we have [ sqrt{n + sqrt{n} + 1} < sqrt{n} + 1. ] By induction, ( a_k(n) < sqrt{n} + 1 ) holds for all ( k ), including ( k = n ). Therefore, [ a_n < sqrt{n} + 1. ] 3. **Combining the Results**: From the above, [ 1 leq frac{a_n}{sqrt{n}} < 1 + frac{1}{sqrt{n}}. ] 4. **Limit Calculation**: Taking the limit as ( n ) approaches infinity, [ lim_{n to infty} left(1 + frac{1}{sqrt{n}}right) = 1. ] Therefore, by the Squeeze Theorem, [ lim_{n to infty} frac{a_n}{sqrt{n}} = 1. ] # Conclusion: [ boxed{1} ]

answer:To solve for the intersection M cap N, we first need to understand the sets M and N individually. For set M, we have: begin{align*} M &= {x|log _{2}x > 1} &= {x|log _{2}x > log _{2}2} &= {x|x > 2} end{align*} This means M includes all x values greater than 2. For set N, we analyze the inequality: begin{align*} N &= {x|frac{x+3}{x-3} < 0} &= {x|(x+3)(x-3) < 0} &= {x|-3 < x < 3} end{align*} This means N includes all x values between -3 and 3, but not including -3 and 3 themselves. To find the intersection M cap N, we combine the conditions for M and N: begin{align*} M cap N &= {x|x > 2} cap {x|-3 < x < 3} &= {x|2 < x < 3} end{align*} This means the intersection of M and N includes all x values that are greater than 2 and less than 3. Therefore, the answer is boxed{B}, which corresponds to the interval (2,3).

answer:1. **Calculating the cube size**: Assume the side length of the cube is (s = 2). 2. **Selecting the tetrahedron vertices**: Given the vertices form a regular tetrahedron along the longest cube diagonals, the vertices of the cube will be ((0,0,0)), ((2,2,2)), ((0,2,2)), and ((2,0,0)). 3. **Tetrahedron side length calculation**: The distance between two vertices, say ((0,0,0)) and ((2,2,2)), is: [ sqrt{(2-0)^2 + (2-0)^2 + (2-0)^2} = 2sqrt{3} ] Thus, the side length of the tetrahedron is (2sqrt{3}). 4. **Computing the tetrahedron's surface area**: [ A = sqrt{3} a^2 = sqrt{3} (2sqrt{3})^2 = sqrt{3} times 12 = 12sqrt{3} ] 5. **Computing the cube's surface area**: [ S = 6s^2 = 6 times (2^2) = 6 times 4 = 24 ] 6. **Ratio of surface areas**: [ frac{S}{A} = frac{24}{12sqrt{3}} = frac{2}{sqrt{3}} = frac{2sqrt{3}}{3} ] The ratio of the surface area of the cube to the surface area of the tetrahedron is (frac{2sqrt{3}{3}}). boxed{The final correct answer is A) (frac{2sqrt{3}}{3}).}

answer:Since line l₀: x - y + 1 = 0, line l₁: ax - 2y + 1 = 0 is perpendicular to l₀, and line l₂: x + by + 3 = 0 is parallel to l₀, we can utilize the relationships between the slopes of parallel and perpendicular lines. 1. First, find the slope of line l₀ by rearranging the equation into slope-intercept form (y = mx + b): y = x + 1, so the slope is 1. 2. As line l₁ is perpendicular to line l₀, the product of their slopes should be equal to -1. The slope of line l₁ can be found by rearranging the equation: y = (a/2)x + 1/2. Therefore, 1 times frac{a}{2} = -1. Solving for 'a', we get a = -2. 3. Since line l₂ is parallel to line l₀, their slopes should be equal. The slope of line l₂ can be found by rearranging the equation: y = (-1/b)x - 3/b. Therefore, -1/b = 1. Solving for 'b', we get b = -1. 4. Finally, compute the value of a + b: -2 + (-1) = -3. Therefore, the value of a + b is boxed{-3}.