\\\\(\\underset{x\ o 0}{\\lim}x\\left[\\dfrac{4}{x}\\right]=A\\\\)\\\\(\\{x\\} = x - [x]\\\\) So, \\\\([x] = x-2\\{x\\}\\\\)\\\\(\\underset{x\ o 0}{\\lim} x \\left(\\dfrac{4}{x} = \\left\\{\\dfrac{4}{x}\\right\\}\\right) = A\\\\)\\\\(\\underset{x\ o 0}{\\lim} 4-x\\left\\{\\dfrac{4}{x}\\right\\} = A\\\\)\\\\(4-0 = A\\\\)\\\\(A = 4\\\\)Checking optionA) \\\\(x = \\sqrt{A} = \\sqrt{4} = 2\\\\) continuousB) \\\\(\\sqrt{A+1} = \\sqrt{5} = \\\\) discontinuousC) \\\\(\\sqrt{A+5} = \\sqrt{9} = 3=\\\\) continuousD) \\\\(\\sqrt{A + 25} = \\sqrt{25} = 5\\\\) continuous