What is the squre root of 6506?

Updated: 4/28/2022
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Q: What is the squre root of 6506?
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What is the limit as x approaches 0 of 4x minus x2 divided by 2 minus square root x?

4x-x^2 \ 2-square root 2 multiply 2+squre root x\ 2+squreroot x = 4x-x^2 (2+squre root x) \ 4_x = x(4-x) (2+squre root x) \ 4-x we will cancel the (4-x) so it will be x(2+squre root x) = (0) (2+squre root 0) =0

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