Question

1. $lim_{x

ightarrow5}(3x^{2}-4x - 1)$

1. $lim_{x

ightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}$

Explanation:

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

For $\lim_{x
ightarrow5}(3x^{2}-4x - 1)$, 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

ightarrow - 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, $\lim_{x
ightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}=\lim_{x
ightarrow - 2}\frac{(x + 2)^{2}}{x^{2}}$

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 $\lim_{x
ightarrow5}(3x^{2}-4x - 1)=54$; for $\lim_{x
ightarrow - 2}\frac{x^{2}+4x + 4}{x^{2}}=0$