Ensino Superior ⇒ Calculo 1- LIMITES Tópico resolvido

[Rafalela]

lim [tex3]\left(\frac{1+x}{2x}\right)^2[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\left(\frac{1+x}{2x}\right)^2[/tex3]

X [tex3]\rightarrow+\infty [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\rightarrow+\infty [/tex3]

[rcompany]

[tex3]
\lim\limits_{x\to+\infty}\left(\frac{1+x}{2x}\right)^2=\lim\limits_{x\to+\infty}\dfrac{1+2x+x^2}{4x^2}=\lim_{x\to+\infty}\dfrac{\dfrac{1}{x^2}+\dfrac{2}{x}+1}{4}=\dfrac{1}{4}\\
\text{já que }\lim\limits_{x\to+\infty}\frac{1}{x^2}=0\text{ e }\lim\limits_{x\to+\infty}\frac{1}{x}=0

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\lim\limits_{x\to+\infty}\left(\frac{1+x}{2x}\right)^2=\lim\limits_{x\to+\infty}\dfrac{1+2x+x^2}{4x^2}=\lim_{x\to+\infty}\dfrac{\dfrac{1}{x^2}+\dfrac{2}{x}+1}{4}=\dfrac{1}{4}\\
\text{já que }\lim\limits_{x\to+\infty}\frac{1}{x^2}=0\text{ e }\lim\limits_{x\to+\infty}\frac{1}{x}=0

[/tex3]