Ensino Superior ⇒ Cálculo desse limite Tópico resolvido

[FIFA2008]

Podem me ajudar a resolver esse limite, por favor?

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[Cardoso1979]

Observe

Uma solução:

[tex3]\lim_{x \rightarrow +\infty}\left[\frac{2(x-1)^2}{\sqrt{4x^2+2x+1}}-x\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}\left[\frac{2(x-1)^2-x\sqrt{4x^2+2x+1}}{\sqrt{4x^2+2x+1}}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\left[\frac{4(x-1)^2-x^2.(4x^2+2x+1)}{(\sqrt{4x^2+2x+1}).[2(x-1)^2+x\sqrt{4x^2+2x+1}}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\left[\frac{\cancel{4x^4}-16x^3+24x^2-16x+4-\cancel{4x^4}-2x^3-x^2}{(2x^2-4x+2).(\sqrt{4x^2+2x+1})+x.(4x^2+2x+1)}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\left[\frac{-18x^3+23x^2-16x+4}{(2x^2-4x+2).\sqrt{x^2.\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}+4x^3+2x^2+x}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\left[\frac{-18x^3+23x^2-16x+4}{(2x^3-4x^2+2x).\sqrt{\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}+x^3.\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\left[\frac{x^3.\left(-18+\frac{23}{x}-\frac{16}{x^2}+\frac{4}{x^3}\right)}{x^3\left(2-\frac{4}{x}+\frac{2}{x^2}\right).\sqrt{\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}+x^3.\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}\right]=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\frac{\cancel{x^3}.\left(-18+\frac{23}{x}-\frac{16}{x^2}+\frac{4}{x^3}\right)}{\cancel{x^3}.\left[\left(2-\frac{4}{x}+\frac{2}{x^2}\right).\sqrt{\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}+4+\frac{2}{x}+\frac{1}{x^2}\right]}=[/tex3]

[tex3]\lim_{x \rightarrow +\infty}
\frac{-18+\frac{23}{x}-\frac{16}{x^2}+\frac{4}{x^3}}{\left[\left(2-\frac{4}{x}+\frac{2}{x^2}\right).\sqrt{\left(4+\frac{2}{x}+\frac{1}{x^2}\right)}+4+\frac{2}{x}+\frac{1}{x^2}\right]}=\frac{-18+0-0+0}{(2-0+0).
\sqrt{4+0+0}+4+0+0}=\frac{-18}{2\sqrt{4}+4}=\frac{-18}{2.2+4}=-\frac{18}{8}=-\frac{9}{4}[/tex3]

Portanto, [tex3]\lim_{x \rightarrow +\infty}\left[\frac{2(x-1)^2}{\sqrt{4x^2+2x+1}}-x\right]=-\frac{9}{4}[/tex3]






Bons estudos!