Find the Domain square root of (sin(x))/x

Problem

√(,sin(x)/x)

Solution

1. Identify the constraints for the domain of the function. For a square root function √(,ƒ(x)) the expression inside must be non-negative: sin(x)/x≥0 Additionally, the denominator cannot be zero, so x≠0

2. Analyze the behavior of the function at the point of discontinuity x=0 Since (lim_x→0)(sin(x)/x)=1 which is greater than zero, the function is defined in a neighborhood around x=0 excluding x=0 itself.

3. Determine the intervals where the numerator and denominator have the same sign. The expression sin(x)/x is positive when sin(x) and x are both positive or both negative.

4. Solve for the case where x>0 We require sin(x)≥0 This occurs on the intervals (0,π]∪[2*π,3*π]∪[4*π,5*π]… which can be written as x∈[2*k*π,(2*k+1)*π] for k∈{0,1,2,…} excluding x=0

5. Solve for the case where x<0 We require sin(x)≤0 This occurs on the intervals [−π,0)∪[−3*π,−2*π]∪[−5*π,−4*π]… which can be written as x∈[(2*k−1)*π,2*k*π] for k∈{0,−1,−2,…} excluding x=0

6. Combine the intervals into a general form. The condition sin(x)/x≥0 is satisfied when x∈(⋃_k∈ℤ∖{0}^)(2*k*π,(2*k+1)*π) if we consider the sign of x relative to the quadrants of the sine function. Specifically, the domain is the union of (−π,0)∪(0,π] and all intervals where x and sin(x) share a sign.

Final Answer

Domain:x∈(⋃_k=1^∞)(2*k*π,(2*k+1)*π)∪(⋃_k=1^∞)(−(2*k+1)*π,−2*k*π)∪(−π,0)∪(0,π]

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