Question

Question #f8576

Answer 1

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