lim x tends to infinity (x-3)^5/(x-5)^5

The answer is one.Because take x common on numeraton and denomeraton and cancel our.we get

(1-3/x)^5/(1-5/x)^5.If you apply limit x tend to infinity 3/x and 5/x become zero.Then you get one.

by L'' hospital rule- differentiate both numerator and denominator simultaneously five times and answer will be 1/1 = 1

lim x tends infenite 5(x-3)^4 / 5(x-5)^4   (infenie/infenite  form) so derivative 5 time to apply limit we get the answer 1

by l''hopitals rule''

x⁵(1-3/x)⁵-x⁵(1-5/x)⁵

cancel x⁵  and 

apply lim 

lim  (x-3)^5/(x-5)^5
x->inf

=e ^ [lim 10/x-5 ]         applying lim [1+ f(x)]^g(x)= e^ lim f(x).g(x)
          x->inf                                x->a                             x->a

=e ^10/x  /  1-5/x

=e^ 0/1

=1 (ANS)