Question

lim (3-x³)⁴-16/x³-1, quando x tende a 1​

Answer 1

Resposta:

lim  [3-x³]⁴-16/(x³-1)

x-->1

lim  {[3-x³]²-4}{[3-x³]²+4}/(x³-1)

x-->1

lim  {[3-x³]-2}{[3-x³]+2}{[3-x³]²+4}/(x³-1)

x-->1

lim  {1-x³}{[3-x³]+2}{[3-x³]²+4}/(x³-1)

x-->1

lim  -{x³-1}{[3-x³]+2}{[3-x³]²+4}/(x³-1)

x-->1

lim  -{[3-x³]+2}{[3-x³]²+4}

x-->1

=-{[3-1³]+2}{[3-1³]²+4}

=-{2+2}*{4+4}

=-4*8=-32 é a resposta

Answer 2

Olá, siga a explicação:

$\mathrm {\displaystyle \lim_{x \to 1} \left ( \dfrac{ \left ( 3-x^3 \right ) ^4 - 16 }{x^3 -1} \right ) }$

Expandi:

$\mathrm { \dfrac{ \left ( 3-x^3 \right )^4 -16 }{x^3 -1} : ~ \dfrac{x^{11} + x^{10} + x^9 - 11x^8 - 11x^7 - 11x^6 + 43x^5 + 43x^4 + 43x^3- 65x^2 - 65x - 65}{x^2 + x +1} }$$\mathrm { \displaystyle \lim_{x \to 1} \left ( \dfrac{x^{11} + x^{10} + x^9- 11x^8-11x^7-11x^6+43x^5+43x^4+43x^3-65x^2-65x -65}{x^2+x+1} \right ) }$$\mathrm { \dfrac{1^{11} + 1^{10} + 1^9 -11 \cdot 1^8 -11 \cdot 1^7 -11 \cdot 1^6 + 43 \cdot 1^5 + 43 \cdot 1^4 + 43 \cdot 1^3 - 65 \cdot 1^2 - 65 \cdot 1 - 65}{1^2+ 1 +1 } }$Simplifica:

$\mathrm { \dfrac{1^{11} + 1^{10} + 1^9 -11 \cdot 1^8 -11 \cdot 1^7 -11 \cdot 1^6 + 43 \cdot 1^5 + 43 \cdot 1^4 + 43 \cdot 1^3 - 65 \cdot 1^2 - 65 \cdot 1 - 65}{3} }$∴ $\boxed { \sf -32}$

• Att. MatiasHP

⇒ Bons Estudos! シ