Question

lim de (4y²+y-3)/(2y+3)(3y+4) com y tendendo ao infinito???

Answer 1

$L=\underset{y\to \infty}{\mathrm{\ell im}}~\dfrac{4y^2+y-3}{(2y+3)(3y+4)}\\\\\\ =\underset{y\to \infty}{\mathrm{\ell im}}~\dfrac{y^2\cdot \left(4+\frac{1}{y}-\frac{3}{y^2}\right)}{y\cdot \left(2+\frac{3}{y} \right )\cdot y\cdot \left(3+\frac{4}{y}\right)}\\\\\\ =\underset{y\to \infty}{\mathrm{\ell im}}~\dfrac{\diagup\!\!\!\!\! y^2\cdot \left(4+\frac{1}{y}-\frac{3}{y^2}\right)}{\diagup\!\!\!\!\! y^2\cdot \left(2+\frac{3}{y} \right )\left(3+\frac{4}{y}\right)}\\\\\\ =\underset{y\to \infty}{\mathrm{\ell im}}~\dfrac{4+\frac{1}{y}-\frac{3}{y^2}}{\left(2+\frac{3}{y} \right )\left(3+\frac{4}{y}\right)}\\\\ \vdots\\\\ =\dfrac{4+0-0}{(2+0)\cdot(3+0)}$

$=\dfrac{4+0-0}{(2+0)\cdot(3+0)}\\\\\\ =\dfrac{4}{2\cdot 3}\\\\\\ =\dfrac{\diagup\!\!\!\! 2\cdot 2}{\diagup\!\!\!\! 2\cdot 3}\\\\\\ =\dfrac{2}{3}\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{y\to \infty}{\mathrm{\ell im}}~\dfrac{4y^2+y-3}{(2y+3)(3y+4)}=\dfrac{2}{3} \end{array}}$


Bons estudos! :-)

Answer 2

lim(4y^2+y-3)/((2y+3)(3y+4)) quando y->+infinito Aplicando a distributiva no denominador: lim(4y^2+y-3)/(6y^2+17y+12) quando y->+infinito Colocando y^2 (termo de maior grau no numerador) e y^2 (termo de maior grau no denominador) em evidência: lim y^2 (4+1/y-3/y^2)/(y^2( (6+17/y+12/y^2)) quando y->+infinito Todas as frações,neste caso,tem lim=0.Então,podemos desconsiderar todas elas para o cálculo,sobrando: lim 4y^2/(6y^2) quando y->+infinito =lim 4/6 quando y->+infinito =4/6 (simplificando por 2)=2/3