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
