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

10 years ago

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.

10 years ago
Applying L''hospital Rule i.e taking five times derivative of numerator and denominator we get limit x tends infinite 120(x-3)/120(x-5) again differentiating numerator and denominator w.r.t x we get 120x/120x which is equals to 1 PLEASE APPROVE MY ANSWER BY CLICKING YES BELOW IF YOU ARE SATISFIED BY IT!!!!!!!!!!!!!! AND BEST OF LUCK FOR YOUR EXAMS!!!!!!!!!!!!!!!!!!!!!!
10 years ago

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

10 years ago

#### 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''

10 years ago

x5(1-3/x)5-x5(1-5/x)5

cancel x5  and

apply lim

 x tends to infinity 3/x and 5/x are 0 and answer becomes1

10 years ago

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)