lim_(x->oo) (2x^2+2x+3)/(x^3+x^2+1)?

1 Answer
Feb 24, 2018

\qquad \qquad \qquad \qquad \qquad \qquad \lim_{ x rarr infty } { 2 x^2 + 2 x + 3 } / { x^3 + x^2 + 1 } \ = \ 0.

Explanation:

"One way to do this is to use the basic (and obvious) fact that:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \lim_{ x rarr infty } 1/x \ = \ 0. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (I)

"We can use this in our example as follows:"

\lim_{ x rarr infty } { 2 x^2 + 2 x + 3 } / { x^3 + x^2 + 1 } \ = \ \lim_{ x rarr infty } { 2 x^2 + 2 x + 3 } / { x^3 + x^2 + 1 } cdot {1/x^3} / {1/x^3}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } { {2 x^2}/{x^3} + {2 x}/{x^3} + {3}/{x^3} } / { {x^3}/{x^3} + {x^2}/{x^3} + {1}/{x^3} }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } { {2}/{x} + {2}/{x^2} + {3}/{x^3} } / { x + {1}/{x} + {1}/{x^3} }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } { 2 ( {1}/{x} ) + 2 ( {1}/{x} )^2 + 3 ( {1}/{x} )^3 } / { x + ( {1}/{x} ) + ( {1}/{x} )^3 }

\qquad "now, using eqn. (I) above, and continuing, we have:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } { 2 ( 0 ) + 2 ( 0 )^2 + 3 ( 0 )^3 } / { x + ( 0 ) + ( 0 )^3 }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } { 0 } / { x }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \lim_{ x rarr infty } 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 0.

"This is our answer."

"Summarizing, we have shown:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \lim_{ x rarr infty } { 2 x^2 + 2 x + 3 } / { x^3 + x^2 + 1 } \ = \ 0.