Calculo 1- LIMITES
Enviado: 22 Out 2021, 10:50
lim [tex3]\left(\frac{1+x}{2x}\right)^2[/tex3]
X [tex3]\rightarrow+\infty [/tex3]
X [tex3]\rightarrow+\infty [/tex3]
[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]