Question

Lim ( 3x + 2x + 5) = x→1
Lim 8 = x→4 *
Lim ( x +9 ) = x→-9 *
​

Answer 1

Respuesta:

\lim _{x\to \infty \:}\left(\frac{\left(2x+2\right)\left(2x+1\right)}{\left(x+1\right)\left(x+1\right)}\right)=4

Explicación paso a paso:

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

=\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)

=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)

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

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

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

=2\cdot \frac{2}{1}

=4