Question

[tex]\lim _{x\to \infty \:}\left(\frac{\left(2x+2\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}\right)[/tex]
ayuden es para mañana amig@s

Answer 1

Verified answer

Respuesta:

[tex]\lim _{x\to \infty \:}\left(\frac{\left(2x+2\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}\right)[/tex]

[tex]2x+2=2x\left(1+\frac{2}{2x}\right)[/tex]

[tex]=\lim _{x\to \infty \:}\left(\frac{2x\left(1+\frac{2}{2x}\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}\right)[/tex]

[tex]\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)[/tex]

[tex]=2\cdot \lim _{x\to \infty \:}\left(\frac{x\left(1+\frac{2}{2x}\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}\right)[/tex]

ahora a simplificar

[tex]\frac{x\left(1+\frac{2}{2x}\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}=\left(\frac{2x+1}{x+1}\right)[/tex]

[tex]=2\cdot \lim _{x\to \infty \:}\left(\frac{2x+1}{x+1}\right)[/tex]

ahora dividir potencia del denominador

[tex]=2\cdot \lim _{x\to \infty \:}\left(\frac{2+\frac{1}{x}}{1+\frac{1}{x}}\right)[/tex]

[tex]\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0[/tex]

forma  indeterminada

[tex]=2\cdot \frac{\lim _{x\to \infty \:}\left(2+\frac{1}{x}\right)}{\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)}[/tex]

[tex]\lim _{x\to \infty \:}\left(2+\frac{1}{x}\right)=2[/tex]

[tex]\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)=1[/tex]

dividir y multiplicar

[tex]=2\cdot \frac{2}{1}[/tex]

resultado

=4