Question

Questão de limites: Me digam o que eu fiz de errado na solução da direita pfvr

Answer 1

$\large\boxed{\begin{array}{l}\sf Determine\,o\,valor\,do\,limite\,a\,seguir:\\\displaystyle\sf\lim_{ x \to 3}\dfrac{x^3-27}{x-3}\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm Recordando\,a\,fatorac_{\!\!,}\tilde ao\,da\,diferenc_{\!\!,}a\,de\,cubos\\\rm temos\,que\, a^3-b^3=(a-b)\cdot(a^2+ab+b^2).\\\rm Isto\,\acute e\, x^3-27=x^3-3^3=(x-3)(x^2+3x+9)\end{array}}$

$\large\boxed{\begin{array}{l}\rm substituindo\,no\,limite\,temos:\\\displaystyle\rm\lim_{ x \to 3}\dfrac{\diagup\!\!\!\!(x-\diagup\!\!\!\!3)\cdot(x^2+3x+9)}{\diagup\!\!\!\!\!(x-\diagup\!\!\!\!\!3)}\\\\\displaystyle\rm\lim_{x \to 3}x^2+3x+9=3^2+3\cdot3+9=27\\\rm nota: x^3-27\ne x^2(x-3)\\\rm pois\, x^2\cdot(x-3)=x^3-3x^2\ne x^3-27.\end{array}}$