Given
[tex]f(x) = \begin{cases} -x^2 - 1 & \text{if }x \neq 5 \\ -3 & \text{if }x=5 \end{cases}[/tex]
we have
[tex]\displaystyle \lim_{x\to5} f(x) = \lim_{x\to5} (-x^2-1) = -5^2-1 = \boxed{-26}[/tex]
since we are approaching [tex]x=5[/tex], so effectively [tex]x\neq 5[/tex] and we use the definition of [tex]f(x)[/tex] under that condition.
