# Question f8576

Feb 21, 2018

1

#### Explanation:

this is equivalent to

$\lim x \to \infty$((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
$\left(\frac{x + 1}{{x}^{2} + 1}\right)$

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

$\therefore \frac{{x}^{2} \left(\frac{1}{x} + \frac{1}{x} ^ 2\right)}{{x}^{2} \left(1 + \frac{1}{x} ^ 2\right)}$

Cancel the like term ${x}^{2}$ to get

$\frac{\frac{1}{x} + \frac{1}{x} ^ 2}{1 + \frac{1}{x} ^ 2}$
as x approaches infinity, the numerator will become 0 as $\frac{1}{x}$ and $\frac{1}{x} ^ 2$ approach 0 as x approaches infinity and likewise the denominator will become 1, therefore the expression becomes $\frac{0}{1} = 0$

therefore the Power is $0$

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

$\left(\frac{{x}^{3} + 5}{{x}^{2} + 2}\right)$
$=$
(x^2(x + 5/x^2))/(x^2(1 + 2/x^2)#

on cancelling the like term x^2 =
$\frac{x + \frac{5}{x} ^ 2}{1 + \frac{2}{x} ^ 2}$
and as anything divided by infinity is 0, the equation becomes
$\frac{x + 0}{1 + 0}$
therefore the whole equation is
$\lim x \rightarrow \infty \left({x}^{0}\right)$ which is
equal to 1