Question

Tentukan hasil dari:
[tex]\displaystyle\lim_{n \to -2} \frac{ {n}^{3} + {n}^{2} - 8n - 12}{ {n}^{2} + 4n + 4} [/tex]
#kls12​

Answer 1

Verified answer

[tex]\displaystyle\lim_{n \to -2} \frac{ {n}^{3} + {n}^{2} - 8n - 12}{ {n}^{2} + 4n + 4}[/tex]

= n³ + 2n² - n² - 2n - 6n - 12/(n + 2)²

= n²( n + 2 ) - n( n + 2 ) - 6( n + 2 )

= (n + 2)(n² - n - 6)/(n + 2)²

= n² + (2 - 3)n - 6/n + 2

= n² + 1n - 6/n + 2

= n² + n - 6/n + 2

= (n + 2)(n - 3)/(n + 2)

= n - 3

= -2 - 3

= -(2 + 3)

= -5