evaluate the limit, if it exists. (if an answer does not exist, enter dne.)\n lim_{x\rightarrow…

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}+4x + 4=(x + 2)^{2}$ by the perfect - square formula $(a + b)^{2}=a^{2}+2ab + b^{2}$ where $a=x$ and $b = 2$. The denominator $x^{4}-16=(x^{2}+4)(x^{2}-4)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x^{2}$ and $b = 4$. Then $x^{2}-4=(x + 2)(x - 2)$. So $x^{4}-16=(x^{2}+4)(x + 2)(x - 2)$.

Step2: Simplify the rational function

The original limit $\lim_{x\rightarrow - 2}\frac{x^{2}+4x + 4}{x^{4}-16}=\lim_{x\rightarrow - 2}\frac{(x + 2)^{2}}{(x^{2}+4)(x + 2)(x - 2)}$. Cancel out the common factor $(x + 2)$ (since $x\neq - 2$ when taking the limit), we get $\lim_{x\rightarrow - 2}\frac{x + 2}{(x^{2}+4)(x - 2)}$.

Step3: Substitute $x=-2$ into the simplified function

Substitute $x=-2$ into $\frac{x + 2}{(x^{2}+4)(x - 2)}$. When $x=-2$, the numerator is $-2 + 2=0$, and the denominator is $((-2)^{2}+4)(-2 - 2)=(4 + 4)(-4)=-32$.

Answer:

$0$