4. $lim_{x\rightarrow5}(3x^{2}-4x - 1)$\n6. $lim_{x\rightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}$

Explanation:

Step1: Substitute $x = 5$ into $3x^{2}-4x - 1$

For $\lim_{x\rightarrow5}(3x^{2}-4x - 1)$, we use the direct - substitution property of limits for polynomial functions. Substitute $x = 5$ into the function $y=3x^{2}-4x - 1$. [ \begin{align*} 3x^{2}-4x - 1&=3\times5^{2}-4\times5 - 1\ &=3\times25-20 - 1\ &=75-20 - 1\ &=54 \end{align*} ]

Step2: Simplify the numerator of $\frac{x^{2}+4x + 4}{x^{2}}$ for $\lim_{x\rightarrow - 2}$

Factor the numerator $x^{2}+4x + 4$. Using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, where $a=x$ and $b = 2$, we have $x^{2}+4x + 4=(x + 2)^{2}$. So, $\lim_{x\rightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}=\lim_{x\rightarrow - 2}\frac{(x + 2)^{2}}{x^{2}}$

Step3: Substitute $x=-2$ into $\frac{(x + 2)^{2}}{x^{2}}$

Substitute $x=-2$ into $\frac{(x + 2)^{2}}{x^{2}}$. [ \begin{align*} \frac{(x + 2)^{2}}{x^{2}}&=\frac{(-2 + 2)^{2}}{(-2)^{2}}\ &=\frac{0^{2}}{4}\ &=0 \end{align*} ]

Answer:

For $\lim_{x\rightarrow5}(3x^{2}-4x - 1)=54$; for $\lim_{x\rightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}=0$