6. $lim_{x\rightarrow2}\frac{x^{2}+4x + 4}{x^{2}}$

Explanation:

Step1: Factor the denominator

The denominator $x^{2}+4x + 4=(x + 2)^{2}$ by the perfect - square formula $(a + b)^{2}=a^{2}+2ab + b^{2}$ where $a=x$ and $b = 2$. So we have $\lim_{x\rightarrow - 2}\frac{x^{2}}{(x + 2)^{2}}$.

Step2: Analyze the limit

As $x\rightarrow - 2$, the numerator $x^{2}\rightarrow(-2)^{2}=4$, and the denominator $(x + 2)^{2}\rightarrow0^{+}$ (since $(x + 2)^{2}\geq0$ for all real $x$ and when $x\rightarrow - 2$, $(x + 2)^{2}$ approaches $0$ from the positive side).

Answer:

$\infty$