Question

Questão de limite
lim (x^100+x^99)/(x^101-x^100 )
(x→∞-)⁡
limite de x^100 + x^99 divido por x^101 - x^100 , x tendendo a infinito pela esquerda.  
Por favor...

Answer 1

$\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}+x^{99}}{x^{101}-x^{100}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}+x^{100-1}}{x^{101}-x^{101-1}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}+x^{100}\cdot x^{-1}}{x^{101}-x^{101}\cdot x^{-1}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}\cdot (1+x^{-1})}{x^{101}\cdot (1-x^{-1})}$

$=\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}\cdot \left(1+\frac{1}{x} \right )}{x^{100+1}\cdot \left(1-\frac{1}{x} \right )}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}\cdot \left(1+\frac{1}{x} \right )}{x^{100}\cdot x\cdot \left(1-\frac{1}{x} \right )}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{1+\frac{1}{x}}{x\cdot \left(1-\frac{1}{x} \right )}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{1}{x}\cdot \dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~f(x)\cdot g(x)~~~~~~\mathbf{(i)}$


onde

$f(x)=\dfrac{1}{x}~~\text{ e }~~g(x)=\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}$

_____________

Sabemos que

$\underset{x\to -\infty}{\mathrm{\ell im}}~f(x)\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{1}{x}=0~~~~~~\mathbf{(ii)}$


e que

$\underset{x\to -\infty}{\mathrm{\ell im}}~g(x)\\\\\\ =\underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{1+\frac{1}{x}}{1-\frac{1}{x}}\\\\\\ \vdots\\\\ =\dfrac{1-0}{1+0}\\\\\\ =\dfrac{1}{1}\\\\ =1~~~~~~\mathbf{(iii)}$


Como o limite das funções $f$ e $g$ existem, então o limite $\mathbf{(i)}$ fica

$=\underset{x\to -\infty}{\mathrm{\ell im}}~f(x)\cdot \underset{x\to -\infty}{\mathrm{\ell im}}~g(x)\\\\ =0\cdot 1\\\\ =0\\\\\\ \therefore~~\boxed{\begin{array}{c} \underset{x\to -\infty}{\mathrm{\ell im}}~\dfrac{x^{100}+x^{99}}{x^{101}-x^{100}}=0 \end{array}}$


Bons estudos! :-)