Należy obliczyć granicę: 

$\underset{x\to -\frac{4}{3}}{\lim }\frac{3x^3+4x^2-6x-8}{9x^2-16}$  

$9x^2-16\ne 0$

$\left(3x-4\right)\left(3x+4\right)\ne 0$

$x\ne \frac{4}{3}\wedge x\ne -\frac{4}{3}$    

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Zauważmy, że:

$3x^3+4x^2-6x-8=x^2\left(3x+4\right)-2\left(3x+4\right)=\left(x^2-2\right)\left(3x+4\right)$  

oraz

$9x^2-16=\left(3x-4\right)\left(3x+4\right)$  

zatem dostajemy:

$\underset{x\to -\frac{4}{3}}{\lim }\frac{3x^3+4x^2-6x-8}{9x^2-16}=\underset{x\to -\frac{4}{3}}{\lim }\frac{\left(x^2-2\right)\left(3x+4\right)}{\left(3x-4\right)\left(3x+4\right)}=$  

$=\underset{x\to -\frac{4}{3}}{\lim }\frac{x^2-2}{3x-4}=\frac{\frac{16}{9}-2}{-4-4}=\frac{-\frac{2}{9}}{-8}=-\frac{2}{9}\cdot \left(-\frac{1}{8}\right)=\frac{1}{36}$  

Wiemy, że:

$\frac{1}{36}\approx 0,0277777$  

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Odp:

$0$

$2$

$7$