Answer:
We have the function [tex]f(x)=x^2+5x\; if \;x\leq -1[/tex] and [tex]f(x)=4x^3-x-1\;if\; x>-1[/tex]
Then [tex]lim_{x\rightarrow -1}f(x)[/tex] exist if there is the limit of f (x) approaching with values on the right of -1, [tex]lim_{x\rightarrow -1^{+}}f(x)[/tex], the limit of f (x) approaching with left values of f (x), [tex]lim_{x\rightarrow -1^{-}}f(x)[/tex], and both coincide.
Observe that the limit of f (x) approaching with values on the right of -1 is
[tex]lim_{x\rightarrow -1^{+}}f(x)=lim_{x\rightarrow -1^{+}}4x^3-x-1=4(-1)^3-(-1)-1=-4+1-1=-4[/tex]
and the limit of f (x) approaching with values on the left of -1 is
[tex]lim_{x\rightarrow -1^{-}}f(x)=lim_{x\rightarrow -1^{-}}(x^2+5x)=(-1)^2+5(-1)=1-5=-4[/tex]
Since the two limits coincide, then
[tex]lim_{x\rightarrow-1}f(x)=-4[/tex]
