raimundojr Escreveu:
Mostre que o limite a seguir é igual a 2:
\(\lim_{(x, y)\rightarrow (0, 0)}\frac{x^4+x^2y^2}{x^2\sqrt{(y-1)^2+2y+(x+2)^2-4x-4}-x^2}\)
\(\lim_{(x, y)\rightarrow (0, 0)}\frac{x^4+x^2y^2}{x^2\sqrt{(y-1)^2+2y+(x+2)^2-4x-4}-x^2}\)
vamos lá.
\(\\\\\\ \lim_{(x, y)\rightarrow (0, 0)}\frac{x^4+x^2y^2}{x^2\sqrt{(y-1)^2+2y+(x+2)^2-4x-4}-x^2} \\\\\\ \lim_{(x, y)\rightarrow (0, 0)}\frac{x^4+x^2y^2}{x^2\sqrt{y^{2}-2y+1+2y+x^{2}+4x+4-4x-4}-x^2} \\\\\\ \lim_{(x, y)\rightarrow (0, 0)}\frac{x^4+x^2y^2}{x^2\sqrt{x^{2}+y^{2}+1}-x^2} \\\\\\ \lim_{(x, y)\rightarrow (0, 0)}\frac{x^{2}(x^2+y^2)}{x^2(\sqrt{x^{2}+y^{2}+1}-1)} \\\\\\ \lim_{(x, y)\rightarrow (0, 0)}\frac{x^2+y^2}{\sqrt{x^{2}+y^{2}+1}-1}\)
agora utilizaremos as coordenadas polares : \(x=r*cos\theta\),\(y=r*sen\theta\) e \(r^2=x^2+y^2\)
\(\\\\\\ \lim_{r\rightarrow 0}\frac{r^2}{\sqrt{r^{2}+1}-1} \\\\\\ \lim_{r\rightarrow 0}\frac{r^2(\sqrt{r^{2}+1}+1)}{(\sqrt{r^{2}+1}-1)*(\sqrt{r^{2}+1}+1)} \\\\\\ \lim_{r\rightarrow 0}\frac{r^2(\sqrt{r^{2}+1}+1)}{r^{2}+1-1}=2 \\\\\\\)
att e bons estudos
