Answer:
We are asked to use the property of limit and continuity to show that the function h(x) is continuous at x= -2.
The function h(x) is given by:
[tex]h(x)=\dfrac{x+3}{(x^2-x-1)(x^2+1)}[/tex]
since clearly as we know that Polynomial functions are continuous everywhere so, the term in the numerator is continuous at x= -2.
Also the term in the denominator is continuous at x= -2.
and the function h(x) is defined in the neighbourhood of x= -2 since the denominator is not equal to zero at x= -2.
Also at x= -2 ; the limit of the function h(x) exist .
The limit is given by:
[tex]h(x)= \lim_{x \to -2} \dfrac{x-2}{(x^2-x-1)(x^2+1)}\\\\h(x)=\dfrac{-2-2}{(-2)^2-(-2)-1)((-2)^2+1)}\\\\h(x)=\dfrac{-4}{4+2-1)(4+1)}\\\\h(x)=\dfrac{-4}{5\times 5}\\\\h(x)=\dfrac{-4}{25}[/tex]
Hence, the function is continuous at x= -2.