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Find the limit of the function algebraically. limit as x approaches negative three of quantity x squared minus nine divided by quantity x cubed plus three.

Respuesta :

LammettHash
The limand is continuous at [tex]x=-3[/tex], so you can directly substitute [tex]x=-3[/tex] to get

[tex]\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\frac{(-3)^2-9}{(-3)^3+3}=\dfrac{9-9}{-27+3}=0[/tex]

Did you mean to write [tex]x^3+3^3=x^3+27[/tex] in the denominator by any chance? In that case, you would instead have

[tex]\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\lim_{x\to-3}\frac{(x+3)(x-3)}{(x+3)(x^2-3x+9)}=\lim_{x\to-3}\frac{x-3}{x^2-3x+9}=\frac{-3-3}{(-3)^2-3(-3)+9}=-\frac29[/tex]

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