lim 3/x.(1/5+x - 1/5-x) resposta é -6/25 preciso da resolução!! Obrigado!!
x->0
Soluções para a tarefa
Respondido por
14
Olá
_____________________________________________________________
Caso não consiga visualizar, tente abrir pelo navegador:
https://brainly.com.br/tarefa/8499697
_____________________________________________________________
![\displaystyle \lim_{x \to 0} ~~ \frac{3}{x} \cdot \left.\left(\dfrac{1}{5+x}-\dfrac{1}{5-x}\right) \displaystyle \lim_{x \to 0} ~~ \frac{3}{x} \cdot \left.\left(\dfrac{1}{5+x}-\dfrac{1}{5-x}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B3%7D%7Bx%7D+%5Ccdot+%5Cleft.%5Cleft%28%5Cdfrac%7B1%7D%7B5%2Bx%7D-%5Cdfrac%7B1%7D%7B5-x%7D%5Cright%29+)
![\displaystyle \text{Tira a constante 3 do limite} \\ \\ \\ 3\cdot\lim_{x \to 0} ~~ \frac{1}{x} \cdot \left.\left(\dfrac{1}{5+x}-\dfrac{1}{5-x}\right) \displaystyle \text{Tira a constante 3 do limite} \\ \\ \\ 3\cdot\lim_{x \to 0} ~~ \frac{1}{x} \cdot \left.\left(\dfrac{1}{5+x}-\dfrac{1}{5-x}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Ctext%7BTira+a+constante+3+do+limite%7D+%5C%5C++%5C%5C++%5C%5C++3%5Ccdot%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B1%7D%7Bx%7D+%5Ccdot+%5Cleft.%5Cleft%28%5Cdfrac%7B1%7D%7B5%2Bx%7D-%5Cdfrac%7B1%7D%7B5-x%7D%5Cright%29+)
Multiplicar por
é a mesma coisa que dividir por X, então dividi a segunda expressão do limite por X;
![\displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{1}{5+x} - \frac{1}{5-x} }{x}}\right) \displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{1}{5+x} - \frac{1}{5-x} }{x}}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E++%5Cleft.%5Cleft%28%5Cdfrac%7B+%5Cfrac%7B1%7D%7B5%2Bx%7D+-+%5Cfrac%7B1%7D%7B5-x%7D+%7D%7Bx%7D%7D%5Cright%29+)
Tira o MMC
![\displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{(5-x)-(5+x)}{(5+x)(5-x)} }{x}}\right) \displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{(5-x)-(5+x)}{(5+x)(5-x)} }{x}}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E++%5Cleft.%5Cleft%28%5Cdfrac%7B+%5Cfrac%7B%285-x%29-%285%2Bx%29%7D%7B%285%2Bx%29%285-x%29%7D+%7D%7Bx%7D%7D%5Cright%29)
![\displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{5-x-5-x}{(5+x)(5-x)} }{x}}\right) \displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{5-x-5-x}{(5+x)(5-x)} }{x}}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E++%5Cleft.%5Cleft%28%5Cdfrac%7B+%5Cfrac%7B5-x-5-x%7D%7B%285%2Bx%29%285-x%29%7D+%7D%7Bx%7D%7D%5Cright%29+)
![\displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{-2x}{(5+x)(5-x)} }{x}}\right) \displaystyle 3\cdot \lim_{x \to 0} ~~ \left.\left(\dfrac{ \frac{-2x}{(5+x)(5-x)} }{x}}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E++%5Cleft.%5Cleft%28%5Cdfrac%7B+%5Cfrac%7B-2x%7D%7B%285%2Bx%29%285-x%29%7D+%7D%7Bx%7D%7D%5Cright%29+)
Temos uma divisão de frações, então, divide a primeira, pelo inverso da segunda
![\displaystyle 3\cdot \lim_{x \to 0} ~~ \frac{-2x}{(5+x)\cdot(5-x)}~ \cdot ~ \frac{1}{x} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2{\diagup\!\!\!\!\!x}}{(5+x)\cdot(5-x)}~ \cdot ~ \frac{1}{\diagup\!\!\!\!\!x} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2}{(5+x)\cdot(5-x)} \\ \\ \\ \text{Substitui o valor de de X no limite} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2}{(5+x)\cdot(5-x)}~ =3\cdot \left.\left(\dfrac{-2}{(5+(0))\cdot(5-(0))}\right) \displaystyle 3\cdot \lim_{x \to 0} ~~ \frac{-2x}{(5+x)\cdot(5-x)}~ \cdot ~ \frac{1}{x} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2{\diagup\!\!\!\!\!x}}{(5+x)\cdot(5-x)}~ \cdot ~ \frac{1}{\diagup\!\!\!\!\!x} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2}{(5+x)\cdot(5-x)} \\ \\ \\ \text{Substitui o valor de de X no limite} \\ \\ \\ 3\cdot \lim_{x \to 0} ~~ \frac{-2}{(5+x)\cdot(5-x)}~ =3\cdot \left.\left(\dfrac{-2}{(5+(0))\cdot(5-(0))}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B-2x%7D%7B%285%2Bx%29%5Ccdot%285-x%29%7D%7E+%5Ccdot+%7E+%5Cfrac%7B1%7D%7Bx%7D+%5C%5C+%5C%5C+%5C%5C+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B-2%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21x%7D%7D%7B%285%2Bx%29%5Ccdot%285-x%29%7D%7E+%5Ccdot+%7E+%5Cfrac%7B1%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21x%7D+%5C%5C+%5C%5C+%5C%5C+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B-2%7D%7B%285%2Bx%29%5Ccdot%285-x%29%7D+%5C%5C+%5C%5C+%5C%5C+%5Ctext%7BSubstitui+o+valor+de+de+X+no+limite%7D+%5C%5C+%5C%5C+%5C%5C+3%5Ccdot+%5Clim_%7Bx+%5Cto+0%7D+%7E%7E+%5Cfrac%7B-2%7D%7B%285%2Bx%29%5Ccdot%285-x%29%7D%7E+%3D3%5Ccdot+%5Cleft.%5Cleft%28%5Cdfrac%7B-2%7D%7B%285%2B%280%29%29%5Ccdot%285-%280%29%29%7D%5Cright%29)
![=3\cdot \left.\left(\dfrac{-2}{(5)\cdot(5)}\right)~=3\cdot \left.\left(\dfrac{-2}{25}\right)~= ~\boxed{ -\frac{6}{25} } =3\cdot \left.\left(\dfrac{-2}{(5)\cdot(5)}\right)~=3\cdot \left.\left(\dfrac{-2}{25}\right)~= ~\boxed{ -\frac{6}{25} }](https://tex.z-dn.net/?f=%3D3%5Ccdot+%5Cleft.%5Cleft%28%5Cdfrac%7B-2%7D%7B%285%29%5Ccdot%285%29%7D%5Cright%29%7E%3D3%5Ccdot+%5Cleft.%5Cleft%28%5Cdfrac%7B-2%7D%7B25%7D%5Cright%29%7E%3D+%7E%5Cboxed%7B+-%5Cfrac%7B6%7D%7B25%7D+%7D)
_____________________________________________________________
Caso não consiga visualizar, tente abrir pelo navegador:
https://brainly.com.br/tarefa/8499697
_____________________________________________________________
Multiplicar por
Tira o MMC
Temos uma divisão de frações, então, divide a primeira, pelo inverso da segunda
Perguntas interessantes