If `f(x)` is continuous at `x=5`, where `f(x)=(sqrt(3+sqrt(4+x))-sqrt(6))/(x-5)`, for `x!=5`, then `f(5)=`

If f(x) is continuous at x=0 , where f(x)=(sqrt(1+x)-root3(1+x))/(x) , for x!=0 , then f(0)=

If f(x) is continuous at x=4 , where f(x)=(x^(4)-64x)/(sqrt(x^(2)+9)-5) , for x!=4 , then f(4)=

If f(x) is continuous at x=sqrt(2) , where f(x)=(sqrt(3+2x)-(sqrt(2)+1))/(x^(2)-2) , for x!= sqrt(2) , then f(sqrt(2))=

If f(x) is continuous at x=a, where f(x) = (sqrt(x)-sqrt(a) + sqrt(x-a))/sqrt(x^(2) -a^(2)) , for x ne a , then f(a)=

If f(x) is continuous at x=a, where f(x) = (sqrt(x)-sqrt(a) + sqrt(x-a))/sqrt(x^(2) -a^(2)) , for x ne a , then f(a)=

If f(x) is continuous at x=2a , where f(x)=(sqrt(x)-sqrt(2a)+sqrt(x-2a))/(sqrt(x^(2)-4a^(2))) , for x!=2a , then f(2a)=

f(x)=sqrt(x-4-2sqrt(x-5))

If f(x) is continuous at x=0 , where f(x)=((4-3x)/(4+5x))^(1/x)", for " x!=0 , then f(0)=

Find the range of f(x) = sqrt(x-1)+sqrt(5-x)