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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
by l''hopitals rule''
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
lim (x-3)^5/(x-5)^5x->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)
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