lim(sqrt(4x^4+x^2+1)-(2x^4+3x^2+x)/(x^2+1)),x->infinity

lim(sqrt(4x^4+x^2+1)-(2x^4+3x^2+x)/(x^2+1)),x->infinity 

$\lim \limits_{x\to \infty } { \left( \sqrt{4x^4+x^2+1}-\dfrac{ 2x^4+3x^2+x } { x^2+1 } \right) } = \ ? \qquad \sqrt{4x^4+x^2+1} = s \qquad s^2 = 4x^4+x^2+1\\ $

$\small{ \begin{array}{rcl} && \sqrt{4x^4+x^2+1} - \dfrac{ 2x^4+3x^2+x } { x^2+1 } \\ &=& s -\dfrac{ 2x^4+3x^2+x } { x^2+1 } \\ &=& \dfrac{ (x^2+1)\cdot s - (2x^4+3x^2+x) } {x^2+1} \\ &=& \left( \dfrac{ (x^2+1)\cdot s - (2x^4+3x^2+x) } {x^2+1} \right) \left( \dfrac{ (x^2+1)\cdot s + (2x^4+3x^2+x) } { (x^2+1)\cdot s + (2x^4+3x^2+x) } \right) \\ &=& \dfrac{ (x^2+1)^2\cdot s^2 - (2x^4+3x^2+x)^2 } { (x^2+1)^2\cdot s + (x^2+1)\cdot (2x^4+3x^2+x) } \\ &=& \dfrac{ (x^2+1)^2\cdot (4x^4+x^2+1) - (2x^4+3x^2+x)^2 } { (x^2+1)^2\cdot s + (x^2+1)\cdot (2x^4+3x^2+x) } \\ &=& \dots \\ &=& \dfrac { 4x^8+9x^6+7x^4+3x^2+1-4x^8-12x^6-4x^5-9x^4-6x^3-x^2 } { (x^4+2x^2+1)\cdot s + 2x^6+5x^4+x^3+3x^2+x } \\ &=& \dfrac { -3x^6-4x^5-2x^4-6x^3+2x^2+1 } { (x^4+2x^2+1)\cdot \sqrt{4x^4+x^2+1} + 2x^6+5x^4+x^3+3x^2+x } \\ &=& \dfrac { -3x^6-4x^5-2x^4-6x^3+2x^2+1 } { (x^4+2x^2+1)\cdot x^2 \sqrt{4+ \frac{1}{x^2}+\frac{1}{x^4} } + 2x^6+5x^4+x^3+3x^2+x } ~ | ~ \frac{:x^6}{:x^6}\\ &=& \dfrac { -3-4\frac{1}{x}-2\frac{1}{x^2}-6\frac{1}{x^3}+2\frac{1}{x^4}+\frac{1}{x^6} } { \frac{(x^4+2x^2+1)}{x^4}\cdot \sqrt{4+ \frac{1}{x^2}+\frac{1}{x^4} } + 2+5\frac{1}{x^2}+\frac{1}{x^3}+3\frac{1}{x^4}+\frac{1}{x^5} } \\ &=& \dfrac { -3-4\frac{1}{x}-2\frac{1}{x^2}-6\frac{1}{x^3}+2\frac{1}{x^4}+\frac{1}{x^6} } { (1+2\frac{1}{x^2}+\frac{1}{x^4})\cdot \sqrt{4+ \frac{1}{x^2}+\frac{1}{x^4} } + 2+5\frac{1}{x^2}+\frac{1}{x^3}+3\frac{1}{x^4}+\frac{1}{x^5} } \\ \lim \limits_{x\to \infty } { \left( \sqrt{4x^4+x^2+1}-\dfrac{ 2x^4+3x^2+x } { x^2+1 } \right) } &=&\dfrac { -3 } { (1)\cdot \sqrt{4} + 2} \\ \lim \limits_{x\to \infty } { \left( \sqrt{4x^4+x^2+1}-\dfrac{ 2x^4+3x^2+x } { x^2+1 } \right) } &=&\dfrac { -3 } { 2 + 2} \\ \mathbf{ \lim \limits_{x\to \infty } { \left( \sqrt{4x^4+x^2+1}-\dfrac{ 2x^4+3x^2+x } { x^2+1 } \right) } } & \mathbf{=} & \mathbf{\dfrac{ -3 }{ 4} }\\ \end{array} }$

lim(sqrt(4x^4+x^2+1)-(2x^4+3x^2+x)/(x^2+1)),x->infinity 

lim_(x->infinity) (sqrt(4 x^4+x^2+1)-(2 x^4+3 x^2+x)/(x^2+1)) = -3/4

Thank you Heureka,

I will study your answer :)

Very nice, heureka......!!! I've added this one to my "tool chest".......!!!!

input syntax for this calulator:

lim(sqrt(4x^4+x^2+1)-(2x^4+3x^2+x)/(x^2+1),x=inf)

Thanks Andre :)