Respuesta :

konrad509
[tex]\displaystyle \lim_{x\to 5}2\cdot(-5^2-4)=2\cdot(-29)=-58\Rightarrow \text{B}[/tex]
gmany
[tex]f(x)=-x^2-4;\ g(x)=2x\\\\g(f(x))=g(-x^2-4)=2(-x^2-4)=-2x^2-8\\\\\lim\limits_{x\to5}g(f(x))=\lim\limits_{x\to5}(-2x^2-8)=(*)\\\\\text{substitute the value of x=5}\\\\(*)=-2\cdot5^2-8=-58\to B.[/tex]