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 $y=3x^{2}-4x - 1$.

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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 $\frac{(x + 2)^{2}}{x^{2}}$.

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Answer:

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