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

Khandavalli Satya Srikanth
38 Points
11 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.

TANAYRAJ SINGH CHOUHAN
65 Points
11 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!!!!!!!!!!!!!!!!!!!!!!
Rajendra Sharma
20 Points
11 years ago

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

Raghuvaran varan Chandragiri
37 Points
11 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''

lokesh soni
37 Points
11 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

Akash Kumar Dutta
98 Points
11 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)