Answer :
[tex] \lim_{x \to -2} \frac{x^2+4x+k}{x+2}
[/tex]
We should eliminate (x+2) in the denominator.
[tex]\lim_{x \to -2} \frac{x^2+4x+k}{x+2} \\ \\ \lim_{x \to -2} \frac{x^2+4x+4}{x+2} \\ \\ \lim_{x \to -2} \frac{(x+2)^2}{x+2} \\ \\ \lim_{x \to -2} {(x+2)}=-2+2=0[/tex]
Therefore, k = 4.
We should eliminate (x+2) in the denominator.
[tex]\lim_{x \to -2} \frac{x^2+4x+k}{x+2} \\ \\ \lim_{x \to -2} \frac{x^2+4x+4}{x+2} \\ \\ \lim_{x \to -2} \frac{(x+2)^2}{x+2} \\ \\ \lim_{x \to -2} {(x+2)}=-2+2=0[/tex]
Therefore, k = 4.