Pré-Vestibular ⇒ (U. Uberaba-MG) Limite

[ALDRINMOD]

O valor do limite [tex3]\lim_{x \to+\infty} (\sqrt{x^2+2x+3}-x)[/tex3] é:

a) [tex3]0[/tex3].
b) [tex3]+\infty[/tex3].
c) [tex3]{-}\infty[/tex3].
d) [tex3]2[/tex3].
e) [tex3]1[/tex3].

Resposta

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e

[Natan]

Vamos ver,

[tex3]\lim_{x \to+\infty} (\sqrt{x^2+2x+3}-x)=\lim_{x \to+\infty} (\sqrt{x^2+2x+3}-x.\frac{\sqrt{x^2+2x+3}+x}{\sqrt{x^2+2x+3}+x})=\lim_{x \to+\infty} \frac{2x+3}{\sqrt{x^2+2x+3}+x}[/tex3]

Proseguiremos agora aplicando L'Hospital:

[tex3]\lim_{x \to+\infty} \frac{2x+3}{\sqrt{x^2+2x+3}+x}=\lim_{x \to+\infty} \frac{2}{\frac{1}{2\sqrt{x^2+2x+3}}+1}=\frac{2}{\frac{1}{2\sqrt{\infty}}+1}=\frac{2}{\frac{1}{\infty}+1}=\frac{2}{0+1}=2[/tex3]