Question #f8576

1 Answer
Feb 21, 2018

1

Explanation:

this is equivalent to

lim x -> oo((x^3 + 5)/(x^2 + 2))^(lim x -> oo ((x+1)/(x^2+1))

lets focus on the Power which the expression is raised to
((x+1)/(x^2+1))

Now factorize the Numerator and denominator with the highest power of x in the denominator

therefore ( x^2(1/x +1/x^2 )) /( x^2( 1 +1/x^2))

Cancel the like term x^2 to get

(1/x +1/x^2 )/( 1 +1/x^2)
as x approaches infinity, the numerator will become 0 as 1/x and 1/x^2 approach 0 as x approaches infinity and likewise the denominator will become 1, therefore the expression becomes 0/1 = 0

therefore the Power is 0

Now use the same factoring on the base using the highest power of x in the denominator

((x^3 + 5)/(x^2 + 2))
=
(x^2(x + 5/x^2))/(x^2(1 + 2/x^2)

on cancelling the like term x^2 =
(x + 5/x^2)/(1 + 2/x^2)
and as anything divided by infinity is 0, the equation becomes
(x + 0)/(1 + 0)
therefore the whole equation is
lim x rarr oo (x^0) which is
equal to 1