Respuesta :

Answer with Step-by-step explanation:

We are given that

[tex]g(x)=(x^2-3x-10)(x^2+2x^2-5x+4)[/tex]

x=-1

Using property of limit

[tex]\lim_{x\rightarrow -1}g(x)=\lim_{x\rightarrow -1}(x^2-3x-10)(3x^2-5x+4)[/tex]

=[tex](1+3-10)(3+5+4)[/tex]

=[tex](-6)(12)[/tex]

=[tex]-72[/tex]

[tex]g(-1)=(1+3-10)(1+2+5+4)[/tex]

[tex]g(-1)=(-6)(12)=-72[/tex]

[tex]\lim_{x\rightarrow -1}g(x)=g(-1)[/tex]

Hence, the function is continuous at x=-1