Question

O $\lim_{x \to -\infty} (x + \sqrt{x^2 + 3x + 2})$ é igual a $- \frac{3}{2}$. Qual é o desenvolvimento dessa questão? Como eu faço?

Answer 1

$L=\underset{x\to -\infty}{\mathrm{\ell im}}~\big(x+\sqrt{x^2+3x+2}\big)\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\big(x+\sqrt{x^2+3x+2}\big)\cdot \dfrac{x-\sqrt{x^2+3x+2}}{x-\sqrt{x^2+3x+2}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{\big(x+\sqrt{x^2+3x+2}\big)\cdot \big(x-\sqrt{x^2+3x+2}\big)}{x-\sqrt{x^2+3x+2}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^2-\big(\sqrt{x^2+3x+2}\big)^2}{x-\sqrt{x^2+3x+2}}$

$=\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^2-(x^2+3x+2)}{x-\sqrt{x^2+3x+2}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^2-x^2-3x-2}{x-\sqrt{x^2+3x+2}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x-\sqrt{x^2+3x+2}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x-\sqrt{x^2\cdot \left(1+\frac{3}{x}+\frac{2}{x^2} \right )}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x-\sqrt{x^2}\cdot\sqrt{1+\frac{3}{x}+\frac{2}{x^2}}}$

$=\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x-|x|\cdot\sqrt{1+\frac{3}{x}+\frac{2}{x^2}}}~~~~~~\mathbf{(i)}$


Mas quando $x\to -\infty,$ temos $x<0.$ Portanto,

$|x|=-x$


Então o limite $\mathbf{(i)}$ fica

$=\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x-(-x)\cdot\sqrt{1+\frac{3}{x}+\frac{2}{x^2}}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3x-2}{x+x\cdot\sqrt{1+\frac{3}{x}+\frac{2}{x^2}}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{\diagup\!\!\!\! x\cdot \left(-3-\frac{2}{x} \right )}{\diagup\!\!\!\! x\cdot \left(1+\sqrt{1+\frac{3}{x}+\frac{2}{x^2}} \right )}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{-3-\frac{2}{x}}{1+\sqrt{1+\frac{3}{x}+\frac{2}{x^2}}}\\\\ \vdots\\\\ =\dfrac{-3+0}{1+\sqrt{1+0+0}}\\\\\\ =\dfrac{-3}{1+\sqrt{1}}\\\\\\ =\dfrac{-3}{1+1}\\\\\\ =-\,\dfrac{3}{2}$


$\therefore~~\boxed{\begin{array}{c} \underset{x\to -\infty}{\mathrm{\ell im}}~\big(x+\sqrt{x^2+3x+2}\big)=-\,\dfrac{3}{2}\end{array}}$


Bons estudos! :-)

Answer 2

$Ola' \\ \\ \lim_{x\to -\infty}~x+ \sqrt{ x^{2} +3x+2} ~~--\ \textgreater \ se~substituirmos -\infty~seria~ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~indeterminado \\ Ent\~ao~fazendo:\\ \lim_{x\to-\infty}~ \frac{(x+ \sqrt{ x^{2} +3x+2}) (x- \sqrt{x\²+3x+2}) }{x- \sqrt{x\²+3x+2} } ~~--\ \textgreater \ deselvolvendo \\ \\ \lim_{x\to-\infty} \frac{x\²-( \sqrt{x\²+3x+2})\² }{x- \sqrt{x\²+3x+2} } \\ \\ \lim_{x\to-\infty}~ \frac{-3x-2}{x- \sqrt{x\²+3x+2} }~~--\ \textgreater \ cambiamos~sinal~de~-\infty ~a~+~\infty$

$\lim_{x\to\infty} \frac{+3x-2}{-x- \underbrace{\sqrt{x\²-3x-2}}_{x} } \\ \\ \lim_{x\to\infty} \frac{3x-2}{-2x} \\ \\ \lim_{x\to\infty} \frac{\not x(3- \frac{2}{x}) }{-2\not x} \\ \\ \lim_{x\to\infty} \frac{3- \frac{2}{x} }{-2} \\ \\ \frac{3- \frac{2}{\infty} }{-2} \\ \\ \boxed{- \frac{3}{2}}} \\ \\ \\ \mathbb{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx} \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$