Question

Calcule os seguintes limites:

Answer 1

Resposta:

d)

lim (x-3)/[x*(x-3)]

x-->3

lim 1/x =1/3

x-->3

e)

P(x)=ax²+bx+c=a*(x-x')*(x-x'')   ....a ≠ 0  ..x' e x'' são as raízes

x²-3x-4=0  ==>x'=-1 e x''=4  , a=1

x²-3x-4 =1*(x+1)*(x-4)

Lim (x+1)*(x-4)/[2*(x+1)]

x-->-1

Lim (x-4)/2 =(-1-4)/2 =-5/2

x-->-1

f)

x²+5x-6=0  ..x'=-6  e x''=1   ..a=1

x²+5x-6=1*(x+6)*(x-1)

LIm (x+6)*(x-1)/(x-1)

x-->1

LIm (x+6) =1+6=7

x-->1

Answer 2

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$\tt d)\\\displaystyle\sf\lim_{ x \to 3}\dfrac{x-3}{x^2-3x}\implies\lim_{x \to 3}\dfrac{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!3)}{x\cdot\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!3)}\\\displaystyle\sf\lim_{x \to 3}\dfrac{1}{x}=\dfrac{1}{3}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{ x \to 3}\dfrac{x-3}{x^2-3x}=\dfrac{1}{3}}}}}\checkmark$

$\tt e)\\\displaystyle\sf\lim_{x \to-1}\dfrac{x^2-3x-4}{2x+2}\\\underbrace{\sf x^2-3x-4=x^2+(1-4)x+(1)\cdot(-4)=(x+1)\cdot(x-4)}_{\rm Produto~de~Stevin}\\\displaystyle\sf\lim_{x \to -1}\dfrac{\diagup\!\!\!\!\!\!(x+\diagup\!\!\!\!\!1)\cdot(x-4)}{2\cdot\diagup\!\!\!\!\!(x+\diagup\!\!\!\!\!1)}\implies \lim_{x \to -1}\dfrac{x-4}{2}=\dfrac{-1-4}{2}=-\dfrac{5}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to -1}\dfrac{x^2-3x-4}{2x+2}=-\dfrac{5}{2}}}}}\checkmark$

$\tt f)\\\displaystyle\sf\lim_{x \to 1}\dfrac{x^2+5x-6}{x-1}\\\sf x^2+5x-6=x^2+(6-1)x+(6)\cdot(-1)=(x+6)\cdot(x-1)\\\displaystyle\sf \lim_{ x \to 1}\dfrac{(x+6)\cdot\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!1)}{\diagup\!\!\!\!\!\!(x-\diagup\!\!\!\!\!1)}\implies \lim_{x \to 1}x+6=1+6=7\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to 1}\dfrac{x^2+5x-6}{x-1}=7}}}}\checkmark$