answer：First, I need to determine how many pencils Cindi bought. Since Cindi spent 30 on pencils that cost 0.50 each, I can calculate the number of pencils by dividing the total amount spent by the price per pencil.Next, Marcia bought double the number of pencils that Cindi did. So, I'll multiply the number of pencils Cindi bought by 2 to find out how many pencils Marcia purchased.Donna bought 3 times as many pencils as Marcia. Therefore, I'll multiply the number of pencils Marcia bought by 3 to find out how many pencils Donna purchased.Finally, to find the total number of pencils bought by Donna and Marcia together, I'll add the number of pencils each of them bought.

answer：To determine the value of |x|, I start by understanding the positions of points A and B on the number line. Point A is at 3 + x, and Point B is at 3 - x.The distance between two points on a number line is the absolute difference of their coordinates. Therefore, the distance between A and B is |(3 + x) - (3 - x)|.Simplifying the expression inside the absolute value:(3 + x) - (3 - x) = 3 + x - 3 + x = 2xSo, the distance between A and B is |2x|.Given that this distance is 8, I set up the equation:|2x| = 8To solve for x, I divide both sides by 2:|x| = 4Thus, the absolute value of x is 4.

answer：Alright, so I have this problem where I need to expand ((2x + 3)^8) and express it in terms of powers of ((x + 1)). Then, I have to find the sum of the coefficients (a + a_2 + a_4 + a_6 + a_8). The options given are A: 6562, B: 3281, C: 3280, D: 6560. Hmm, okay.First, let me understand what's being asked. The expression ((2x + 3)^8) is given, and I need to rewrite it as a polynomial in ((x + 1)). That is, instead of having terms like (x^k), I need to express it in terms of ((x + 1)^k). Once that's done, I have to sum the coefficients of the even-powered terms, specifically (a_0, a_2, a_4, a_6, a_8). I remember that when you have a polynomial expressed in terms of ((x + c)), you can use the binomial theorem or substitution to find the coefficients. Maybe I can substitute (y = x + 1), so that (x = y - 1). Then, substitute this into the original expression to express everything in terms of (y). That should give me the expansion in terms of ((x + 1)), which is (y). Let me try that. So, let (y = x + 1), which implies (x = y - 1). Substitute this into ((2x + 3)^8):[(2(y - 1) + 3)^8]Simplify inside the parentheses:[(2y - 2 + 3)^8 = (2y + 1)^8]Okay, so now the expression becomes ((2y + 1)^8). That's a binomial expression, which I can expand using the binomial theorem. The binomial theorem states that:[(a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k]In this case, (a = 2y), (b = 1), and (n = 8). So, expanding this:[(2y + 1)^8 = sum_{k=0}^{8} binom{8}{k} (2y)^{8 - k} (1)^k]Simplify each term:[= sum_{k=0}^{8} binom{8}{k} 2^{8 - k} y^{8 - k}]Alternatively, I can write this as:[= sum_{m=0}^{8} binom{8}{m} 2^{m} y^{m}]Wait, actually, hold on. Let me make sure I get the substitution right. When I set (m = 8 - k), then when (k = 0), (m = 8), and when (k = 8), (m = 0). So, reversing the order, the expansion becomes:[= sum_{m=0}^{8} binom{8}{8 - m} 2^{m} y^{m}]But since (binom{8}{8 - m} = binom{8}{m}), this simplifies to:[= sum_{m=0}^{8} binom{8}{m} 2^{m} y^{m}]So, the expansion is:[(2y + 1)^8 = sum_{m=0}^{8} binom{8}{m} 2^{m} y^{m}]Which means that the coefficients (a_m) in the original problem are (binom{8}{m} 2^{m}). Therefore, the coefficients (a_0, a_1, ..., a_8) are given by:[a_m = binom{8}{m} 2^{m}]So, to find (a + a_2 + a_4 + a_6 + a_8), which is (a_0 + a_2 + a_4 + a_6 + a_8), I need to compute the sum of the even-indexed coefficients in this expansion.I remember that for a polynomial (P(y) = sum_{m=0}^{n} c_m y^m), the sum of the coefficients can be found by evaluating (P(1)), and the alternating sum can be found by evaluating (P(-1)). Specifically:- (P(1) = sum_{m=0}^{n} c_m)- (P(-1) = sum_{m=0}^{n} c_m (-1)^m)If I add these two, I get:[P(1) + P(-1) = sum_{m=0}^{n} c_m (1 + (-1)^m) = 2 sum_{k=0}^{lfloor n/2 rfloor} c_{2k}]Which is twice the sum of the even-indexed coefficients. Similarly, subtracting them gives twice the sum of the odd-indexed coefficients.So, in this case, (P(y) = (2y + 1)^8). Therefore:[P(1) = (2(1) + 1)^8 = 3^8][P(-1) = (2(-1) + 1)^8 = (-2 + 1)^8 = (-1)^8 = 1]Adding these together:[P(1) + P(-1) = 3^8 + 1]Which equals twice the sum of the even-indexed coefficients. Therefore, the sum of the even-indexed coefficients is:[frac{P(1) + P(-1)}{2} = frac{3^8 + 1}{2}]Let me compute (3^8). I know that (3^2 = 9), (3^3 = 27), (3^4 = 81), (3^5 = 243), (3^6 = 729), (3^7 = 2187), and (3^8 = 6561). So, (3^8 = 6561). Therefore:[frac{6561 + 1}{2} = frac{6562}{2} = 3281]So, the sum (a_0 + a_2 + a_4 + a_6 + a_8 = 3281). Looking back at the options, that's option B: 3281.Wait, but hold on a second. Let me double-check my substitution. I set (y = x + 1), so (x = y - 1). Then, substituting into ((2x + 3)^8) gives ((2(y - 1) + 3)^8), which simplifies to ((2y - 2 + 3)^8 = (2y + 1)^8). That seems correct.Then, expanding ((2y + 1)^8) using the binomial theorem, I get the coefficients as (binom{8}{m} 2^m). So, the coefficients (a_m) are indeed (binom{8}{m} 2^m). Calculating (P(1)) and (P(-1)) seems correct as well. (P(1) = 3^8 = 6561), (P(-1) = (-1)^8 = 1). So, adding them gives 6562, and dividing by 2 gives 3281. That seems correct.But wait, let me think again. The question says "find the value of (a + a_2 + a_4 + a_6 + a_8)", which is the sum of the coefficients of ((x + 1)^0, (x + 1)^2, ..., (x + 1)^8). So, in the expansion, these are the even-powered terms, which correspond to the even (m) in the expression (P(y)). Therefore, the method of using (P(1) + P(-1)) is correct because it isolates the even-powered coefficients.Alternatively, another way to think about it is that if I plug in (y = 1), I get the sum of all coefficients, and if I plug in (y = -1), I get the alternating sum. Adding them cancels out the odd terms and doubles the even terms, so dividing by 2 gives the sum of the even coefficients.Therefore, my calculation seems correct. So, the answer should be 3281, which is option B.Wait, but hold on. Let me check if I made a mistake in the substitution step. So, the original expression is ((2x + 3)^8), and we're expressing it in terms of ((x + 1)). So, (x = y - 1), so substituting, we get:[(2(y - 1) + 3)^8 = (2y - 2 + 3)^8 = (2y + 1)^8]Yes, that's correct. So, the expansion is indeed ((2y + 1)^8), and the coefficients are as I calculated.Alternatively, maybe I can compute the sum directly by calculating each (a_{2k}) and adding them up. Let's see:Compute (a_0 = binom{8}{0}2^0 = 1)(a_2 = binom{8}{2}2^2 = 28 * 4 = 112)(a_4 = binom{8}{4}2^4 = 70 * 16 = 1120)(a_6 = binom{8}{6}2^6 = 28 * 64 = 1792)(a_8 = binom{8}{8}2^8 = 1 * 256 = 256)Now, sum these up:1 + 112 = 113113 + 1120 = 12331233 + 1792 = 30253025 + 256 = 3281So, that's the same result as before. So, 3281 is indeed the correct sum.Therefore, the answer is B: 3281.But wait, hold on another thought. The problem says "Expand ((2x + 3)^8 = a + a_1(x + 1) + a_2(x + 1)^2 + ldots + a_8(x + 1)^8)", so the constant term is (a), which is (a_0). So, when I computed (a_0 + a_2 + a_4 + a_6 + a_8), that is exactly the sum of the coefficients of the even powers of ((x + 1)). So, that's correct.Alternatively, another way to approach this is to consider evaluating the polynomial at specific points. For example, if I set (x = 0), then the expression becomes ((2*0 + 3)^8 = 3^8 = 6561). On the other hand, substituting (x = 0) into the expansion gives (a + a_1(1) + a_2(1)^2 + ldots + a_8(1)^8 = a + a_1 + a_2 + ldots + a_8 = 6561). Similarly, if I set (x = -2), then ((2*(-2) + 3)^8 = (-4 + 3)^8 = (-1)^8 = 1). Substituting (x = -2) into the expansion gives (a + a_1(-1) + a_2(1)^2 + ldots + a_8(-1)^8 = a - a_1 + a_2 - a_3 + a_4 - a_5 + a_6 - a_7 + a_8 = 1).So, if I denote (S = a + a_2 + a_4 + a_6 + a_8) and (T = a_1 + a_3 + a_5 + a_7), then from (x = 0), we have (S + T = 6561), and from (x = -2), we have (S - T = 1). Adding these two equations:(2S = 6562) => (S = 3281)Which is the same result as before. So, that's another way to confirm it.Therefore, I'm confident that the correct answer is 3281, which is option B.**Final Answer**boxed{B}

answer：Okay, so I have this system of equations:[begin{cases}x + 2y = k + 2 2x - 3y = 3k - 1end{cases}]And I need to find the value of ( x + 9y ) which is a constant for any value of ( k ). Hmm, interesting. Let me think about how to approach this.First, maybe I can solve the system for ( x ) and ( y ) in terms of ( k ) and then plug those into ( x + 9y ) to see if it simplifies to a constant.So, let's write down the equations again:1. ( x + 2y = k + 2 ) -- Equation (1)2. ( 2x - 3y = 3k - 1 ) -- Equation (2)I can use either substitution or elimination. Let me try elimination because the coefficients look manageable.If I multiply Equation (1) by 2, I can eliminate ( x ):Multiply Equation (1) by 2:( 2x + 4y = 2k + 4 ) -- Equation (3)Now subtract Equation (2) from Equation (3):( (2x + 4y) - (2x - 3y) = (2k + 4) - (3k - 1) )Simplify the left side:( 2x + 4y - 2x + 3y = 7y )Right side:( 2k + 4 - 3k + 1 = (-k) + 5 )So, we have:( 7y = -k + 5 )Therefore, ( y = frac{-k + 5}{7} )Okay, so that's ( y ) in terms of ( k ). Now, let's find ( x ).From Equation (1):( x = k + 2 - 2y )We already have ( y = frac{-k + 5}{7} ), so plug that in:( x = k + 2 - 2 times left( frac{-k + 5}{7} right) )Let me compute that step by step.First, compute ( 2 times left( frac{-k + 5}{7} right) ):( frac{-2k + 10}{7} )So, ( x = k + 2 - frac{-2k + 10}{7} )To combine these terms, let me express ( k + 2 ) as a fraction with denominator 7:( x = frac{7k + 14}{7} - frac{-2k + 10}{7} )Now, subtract the numerators:( (7k + 14) - (-2k + 10) = 7k + 14 + 2k - 10 = 9k + 4 )So, ( x = frac{9k + 4}{7} )Alright, so now we have both ( x ) and ( y ) in terms of ( k ):( x = frac{9k + 4}{7} )( y = frac{-k + 5}{7} )Now, the problem asks for ( x + 9y ). Let's compute that.First, compute ( 9y ):( 9y = 9 times frac{-k + 5}{7} = frac{-9k + 45}{7} )Now, add ( x ):( x + 9y = frac{9k + 4}{7} + frac{-9k + 45}{7} )Combine the numerators:( (9k + 4) + (-9k + 45) = 0k + 49 = 49 )So, ( x + 9y = frac{49}{7} = 7 )Wait, so ( x + 9y = 7 ). That's a constant, as expected. So, regardless of the value of ( k ), ( x + 9y ) is always 7.Let me double-check my steps to make sure I didn't make a mistake.1. Solved the system using elimination, found ( y = frac{-k + 5}{7} ). That seems right.2. Plugged ( y ) back into Equation (1) to find ( x = frac{9k + 4}{7} ). That also seems correct.3. Then computed ( x + 9y ), which gave me 7. Let me verify:Compute ( x + 9y ):( x = frac{9k + 4}{7} )( 9y = 9 times frac{-k + 5}{7} = frac{-9k + 45}{7} )Adding them together:( frac{9k + 4 - 9k + 45}{7} = frac{49}{7} = 7 ). Yep, that's correct.Alternatively, maybe there's another way to see this without solving for ( x ) and ( y ). Let me think.If ( x + 9y ) is a constant, then perhaps we can express ( x + 9y ) as a linear combination of the two equations.Let me denote the two equations as:Equation (1): ( x + 2y = k + 2 )Equation (2): ( 2x - 3y = 3k - 1 )Suppose we find constants ( a ) and ( b ) such that:( a times text{Equation (1)} + b times text{Equation (2)} = x + 9y + c times k + d )But since ( x + 9y ) is a constant, the coefficients of ( k ) should cancel out.Wait, let's try to write ( x + 9y ) as a combination.Let me set up the equations:( a(x + 2y) + b(2x - 3y) = x + 9y + (a + 3b)k + (2a - b) )Wait, maybe not. Let me think again.Wait, actually, if I take a linear combination of the two equations, the right-hand side will involve ( k ). But since ( x + 9y ) is supposed to be a constant, the coefficient of ( k ) in the right-hand side must be zero.So, let's write:( a(x + 2y) + b(2x - 3y) = (a + 2b)x + (2a - 3b)y = x + 9y )So, equate coefficients:For ( x ): ( a + 2b = 1 )For ( y ): ( 2a - 3b = 9 )Now, solve this system for ( a ) and ( b ).Equation (A): ( a + 2b = 1 )Equation (B): ( 2a - 3b = 9 )Let's solve Equation (A) for ( a ):( a = 1 - 2b )Plug into Equation (B):( 2(1 - 2b) - 3b = 9 )Simplify:( 2 - 4b - 3b = 9 )( 2 - 7b = 9 )Subtract 2:( -7b = 7 )Divide by -7:( b = -1 )Now, plug back into ( a = 1 - 2b ):( a = 1 - 2(-1) = 1 + 2 = 3 )So, ( a = 3 ), ( b = -1 )Therefore, the combination is:( 3 times text{Equation (1)} - 1 times text{Equation (2)} = x + 9y )Compute the right-hand side:( 3(k + 2) - 1(3k - 1) = 3k + 6 - 3k + 1 = 7 )So, ( x + 9y = 7 ). That's another way to see it without solving for ( x ) and ( y ).Either way, I get the same result. So, I'm confident that the constant value is 7.**Final Answer**The constant value of ( x + 9y ) is boxed{7}.

answer：Okay, so I have this problem here: given that a and b are both positive numbers, and the function f(x) = 4x³ - a x² - 2b x + 2 has an extreme value at x = 1. I need to find the maximum value of ab. Hmm, okay, let's break this down step by step.First, I remember that extreme values of a function occur where the first derivative is zero or undefined. Since f(x) is a polynomial, its derivative will also be a polynomial, which is defined everywhere. So, the extreme value at x = 1 means that the first derivative of f(x) evaluated at x = 1 is zero. That gives me an equation to work with.Let me compute the first derivative of f(x). The function is f(x) = 4x³ - a x² - 2b x + 2. Taking the derivative term by term:- The derivative of 4x³ is 12x².- The derivative of -a x² is -2a x.- The derivative of -2b x is -2b.- The derivative of the constant term 2 is 0.So, putting it all together, f'(x) = 12x² - 2a x - 2b.Now, since there's an extreme at x = 1, we know that f'(1) = 0. Let's plug x = 1 into the derivative:f'(1) = 12(1)² - 2a(1) - 2b = 0.Simplifying that:12 - 2a - 2b = 0.Let me write that equation down:12 - 2a - 2b = 0.I can simplify this equation by dividing both sides by 2:6 - a - b = 0.Which simplifies to:a + b = 6.So, the sum of a and b is 6. But I need to find the maximum value of ab. Hmm, okay, so this seems like an optimization problem where I need to maximize the product ab given that a + b = 6, with a > 0 and b > 0.I remember that for two positive numbers with a fixed sum, their product is maximized when the numbers are equal. Is that correct? Let me verify.Yes, from the AM-GM inequality, the arithmetic mean is greater than or equal to the geometric mean. So, for a and b,(a + b)/2 ≥ √(ab).Since a + b = 6, this becomes:6/2 ≥ √(ab) => 3 ≥ √(ab).Squaring both sides:9 ≥ ab.So, the maximum value of ab is 9, achieved when a = b = 3. Wait, but hold on a second. Is that the case here? Let me think.In this problem, a and b are positive, so they can be any positive numbers adding up to 6. If I set a = b = 3, then ab = 9, which is the maximum. So, that seems straightforward.But wait, let me make sure that this is indeed the case. Maybe there's something else I need to consider because the function is a cubic, and having an extreme at x = 1 might impose some other conditions on a and b.Let me think about the second derivative to check if x = 1 is a maximum or a minimum. The second derivative of f(x) is f''(x). Let's compute that.We had f'(x) = 12x² - 2a x - 2b. Taking the derivative again:f''(x) = 24x - 2a.Evaluating at x = 1:f''(1) = 24(1) - 2a = 24 - 2a.Now, for x = 1 to be an extreme value, it's already given, but we can check the concavity. If f''(1) > 0, then it's a local minimum; if f''(1) < 0, it's a local maximum. But the problem doesn't specify whether it's a maximum or minimum, just that it's an extreme value. So, I don't think this affects the condition a + b = 6.Therefore, regardless of whether it's a maximum or minimum, the condition is just a + b = 6. So, the maximum of ab is 9 when a = b = 3.Wait, but hold on again. Let me check if a and b can indeed be 3 each. If a = 3 and b = 3, then let's plug back into the original function and see if x = 1 is indeed an extreme.f(x) = 4x³ - 3x² - 6x + 2.Compute f'(x):f'(x) = 12x² - 6x - 6.At x = 1:f'(1) = 12(1) - 6(1) - 6 = 12 - 6 - 6 = 0. So, that's correct.Compute f''(1):f''(x) = 24x - 6.f''(1) = 24(1) - 6 = 18, which is positive, so it's a local minimum. But the problem didn't specify whether it's a maximum or minimum, just that it's an extreme. So, that's fine.Therefore, a = 3, b = 3, ab = 9 is acceptable.But wait, is there any other condition that I might have missed? For example, maybe the function needs to have another extreme or something else? Let me check.The function is a cubic, so it must have two critical points: one local maximum and one local minimum. Since it's given that there's an extreme at x = 1, but we don't know about the other one. So, maybe the other critical point is somewhere else, but that shouldn't affect the condition a + b = 6.Alternatively, perhaps I need to ensure that x = 1 is the only critical point? But no, a cubic function must have two critical points, so x = 1 is one, and there's another one somewhere else.But since the problem only mentions an extreme at x = 1, I think the only condition we have is a + b = 6. So, the maximum of ab is 9.Wait, but let me think again. Maybe I can approach this problem using calculus, treating it as an optimization problem with a constraint.So, we have the constraint a + b = 6, and we need to maximize ab. So, we can set up the problem as maximizing ab with a + b = 6.Let me express b in terms of a: b = 6 - a.Then, ab = a(6 - a) = 6a - a².To find the maximum of this quadratic function, we can take its derivative with respect to a:d/da (6a - a²) = 6 - 2a.Set derivative equal to zero:6 - 2a = 0 => 2a = 6 => a = 3.So, a = 3, then b = 6 - 3 = 3, so ab = 9.Therefore, the maximum value of ab is indeed 9.Wait, but is there any other way this could be approached? Maybe using Lagrange multipliers or something else? Hmm, but since it's a simple constraint, substitution seems sufficient.Alternatively, since ab is a quadratic function in terms of a, it's a downward opening parabola, so its maximum is at the vertex, which is at a = -b/(2a) in the quadratic formula. Wait, in this case, the quadratic is -a² + 6a, so the vertex is at a = -6/(2*(-1)) = 3, which is the same result.So, yeah, that's consistent.But just to make sure, let me consider if a and b can take other values that might give a higher product ab while still satisfying a + b = 6. For example, if a is 4, then b is 2, so ab = 8, which is less than 9. If a is 5, b is 1, ab = 5, which is even less. If a is 2, b is 4, ab = 8. So, indeed, the maximum occurs when a and b are equal.Therefore, I think the maximum value of ab is 9, achieved when a = b = 3.But hold on, is there any possibility that a or b could be zero? The problem states a > 0 and b > 0, so neither can be zero. So, a and b have to be positive numbers adding up to 6. So, 3 and 3 is the midpoint, which gives the maximum product.Alternatively, if a or b were allowed to be zero, then ab could be zero, which is less than 9, so that doesn't affect our maximum.So, I think I've considered all the angles here. The key was recognizing that the condition a + b = 6 comes from the derivative at x = 1, and then using optimization techniques to find the maximum product ab under that constraint.Therefore, the maximum value of ab is 9.**Final Answer**The maximum value of ( ab ) is boxed{9}.

answer：Alright, so I have this limit to compute:[lim _{x rightarrow 0} sqrt{operatorname{arctg} x cdot sin ^{2}left(frac{1}{x}right)+5 cos x}]Hmm, okay. Let me break this down step by step. I remember that when dealing with limits, especially as x approaches a certain point, I should look at each part of the function separately and see how they behave.First, let me rewrite the expression inside the square root to make it clearer:[sqrt{operatorname{arctg} x cdot sin ^{2}left(frac{1}{x}right) + 5 cos x}]So, the function inside the square root is the sum of two terms: one involving arctangent and sine squared, and another involving cosine. Let me analyze each term as x approaches 0.Starting with the first term: (operatorname{arctg} x cdot sin ^{2}left(frac{1}{x}right)).I know that as x approaches 0, (operatorname{arctg} x) behaves like x. Because the Taylor series expansion of arctangent around 0 is x - x^3/3 + x^5/5 - ..., so for small x, arctg x ≈ x. So, that part is straightforward.Now, the other part is (sin ^{2}left(frac{1}{x}right)). Hmm, sine squared of 1/x. As x approaches 0, 1/x approaches infinity. So, sin(1/x) oscillates between -1 and 1, and when squared, it oscillates between 0 and 1. So, sin²(1/x) is always between 0 and 1, regardless of how x approaches 0.Therefore, (sin ^{2}left(frac{1}{x}right)) is bounded between 0 and 1. So, when we multiply it by arctg x, which is approximately x near 0, we get a term that is approximately x times something between 0 and 1. So, as x approaches 0, this term should approach 0, right? Because x is going to 0, and it's multiplied by something that doesn't blow up but stays within 0 and 1.So, the first term, (operatorname{arctg} x cdot sin ^{2}left(frac{1}{x}right)), tends to 0 as x approaches 0.Now, moving on to the second term inside the square root: 5 cos x. As x approaches 0, cos x approaches 1. So, 5 cos x approaches 5 * 1 = 5.Therefore, inside the square root, we have the sum of two terms: one approaching 0 and the other approaching 5. So, the entire expression inside the square root approaches 0 + 5 = 5.Thus, the expression under the square root approaches 5. So, the square root of 5 is just sqrt(5). Therefore, the limit should be sqrt(5).Wait, but hold on. Let me make sure I'm not missing anything. The first term is arctg x times sin²(1/x). I said that arctg x is approximately x, and sin²(1/x) is bounded. So, x times a bounded function is 0. But is that always the case?Yes, because as x approaches 0, x goes to 0, and even though sin²(1/x) oscillates, it doesn't go to infinity or anything; it just stays between 0 and 1. So, multiplying by x, which is approaching 0, the whole term goes to 0.So, the first term is negligible compared to the second term, which approaches 5. Therefore, the expression inside the square root approaches 5, and the square root of 5 is just a constant.Therefore, the limit is sqrt(5). So, I think that's the answer.But just to double-check, let me consider the behavior of the function as x approaches 0 from both sides. Since arctg x is an odd function, but sin²(1/x) is even, so the first term is even because arctg x is odd and sin²(1/x) is even, so their product is odd times even, which is odd. But wait, no, arctg x is odd, sin²(1/x) is even, so their product is odd. Hmm, but when multiplied by each other, it's an odd function. But since we're taking the square root, which is always positive, does that affect anything?Wait, no, because inside the square root, we have the sum of two terms. The first term is odd, but the second term is even because cos x is even. So, the entire expression inside the square root is even because 5 cos x is even, and the first term is odd, but when added to an even function, the result is neither even nor odd necessarily. Wait, but as x approaches 0, whether from the positive or negative side, arctg x approaches 0, and sin²(1/x) is still between 0 and 1 regardless of the direction. So, the first term is approaching 0 from both sides, and the second term is approaching 5 from both sides.Therefore, the expression inside the square root approaches 5 from both sides, so the square root is approaching sqrt(5) regardless of the direction. So, the limit exists and is equal to sqrt(5).Is there any chance that the first term could cause some oscillation that might affect the limit? For example, if the first term oscillated wildly, could it cause the expression inside the square root to oscillate as well?Well, the first term is arctg x times sin²(1/x). As x approaches 0, arctg x approaches 0, and sin²(1/x) oscillates between 0 and 1. So, the first term oscillates between 0 and arctg x, which is approaching 0. So, the first term is oscillating between 0 and something approaching 0. So, the entire expression inside the square root is oscillating between 5 and 5 + something approaching 0. So, it's oscillating between 5 and just above 5, but the limit is still 5 because the oscillations die out as x approaches 0.Wait, actually, no. Because sin²(1/x) is always non-negative, so the first term is non-negative, so the expression inside the square root is always 5 plus something non-negative approaching 0. So, the expression inside the square root is approaching 5 from above, but the square root of a number approaching 5 from above is just approaching sqrt(5). So, even though the expression inside is oscillating, it's oscillating within a neighborhood of 5, and the square root is continuous there, so the limit is sqrt(5).Therefore, I think my initial conclusion is correct.So, to recap:1. Break down the expression inside the square root into two terms.2. Analyze each term as x approaches 0.3. The first term, arctg x times sin²(1/x), approaches 0 because arctg x ~ x, and sin²(1/x) is bounded.4. The second term, 5 cos x, approaches 5.5. Therefore, the expression inside the square root approaches 5.6. The square root of 5 is a constant, so the limit is sqrt(5).I think that covers everything. I don't see any mistakes in this reasoning. So, I can confidently say that the limit is sqrt(5).**Final Answer**The limit is boxed{sqrt{5}}.