Question

evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter dne.)
\\(\lim_{x\to8}(1 + \sqrt3{x})(2 - 6x^{2}+x^{3})\\)

Explanation:

Step1: Apply product - limit law

The product - limit law states that $\lim_{x
ightarrow a}(f(x)g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$. Here, $f(x)=1 + \sqrt[3]{x}$ and $g(x)=2-6x^{2}+x^{3}$. So, $\lim_{x
ightarrow 8}(1 + \sqrt[3]{x})(2-6x^{2}+x^{3})=\lim_{x
ightarrow 8}(1 + \sqrt[3]{x})\cdot\lim_{x
ightarrow 8}(2-6x^{2}+x^{3})$.

Step2: Evaluate $\lim_{x

ightarrow 8}(1 + \sqrt[3]{x})$
Substitute $x = 8$ into $1+\sqrt[3]{x}$. We know that $\sqrt[3]{8}=2$, so $\lim_{x
ightarrow 8}(1 + \sqrt[3]{x})=1+\sqrt[3]{8}=1 + 2=3$.

Step3: Evaluate $\lim_{x

ightarrow 8}(2-6x^{2}+x^{3})$
Use the sum - difference and constant - multiple limit laws. $\lim_{x
ightarrow 8}(2-6x^{2}+x^{3})=\lim_{x
ightarrow 8}2-6\lim_{x
ightarrow 8}x^{2}+\lim_{x
ightarrow 8}x^{3}$. Since $\lim_{x
ightarrow a}c=c$ (where $c$ is a constant), $\lim_{x
ightarrow 8}2 = 2$. Also, $\lim_{x
ightarrow a}x^{n}=a^{n}$, so $6\lim_{x
ightarrow 8}x^{2}=6\times8^{2}=6\times64 = 384$ and $\lim_{x
ightarrow 8}x^{3}=8^{3}=512$. Then $\lim_{x
ightarrow 8}(2-6x^{2}+x^{3})=2-384 + 512=130$.

Step4: Multiply the two limits

Multiply the results from Step 2 and Step 3. $\lim_{x
ightarrow 8}(1 + \sqrt[3]{x})\cdot\lim_{x
ightarrow 8}(2-6x^{2}+x^{3})=3\times130 = 390$.

Answer:

390