Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE). $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-41 x^{5}+x^{6}}}{5 x+5 x^{2}}$ $\newline$ Answer:

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Q. Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE).

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\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-41 x^{5}+x^{6}}}{5 x+5 x^{2}}

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Answer:

Factor out highest power: We are given the limit: $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt[3]{-41x^{5}+x^{6}}}{5x+5x^{2}}\right)$ $\newline$ First, we will factor out the highest power of $x$ in the numerator and denominator to simplify the expression.
Factor out $x^{6}$ : In the numerator, the highest power of $x$ is $x^{6}$ , so we factor $x^{6}$ out of the cube root. $\newline$ $\sqrt[3]{-41x^{5}+x^{6}} = \sqrt[3]{x^{6}(-41/x+1)}$
Factor out $x^{2}$ : In the denominator, the highest power of $x$ is $x^{2}$ , so we factor $x^{2}$ out. $\newline$ $5x+5x^{2} = 5x(1+x)$
Rewrite with factored terms: Now we rewrite the limit with the factored terms: \lim_{x \to \infty}\left(\sqrt[\(3]{x^{ $6$ }(-\frac{ $41$ }{x}+ $1$ )}\right)/\left( $5$ x( $1$ +x)\right)
Simplify cube root: We can simplify the cube root of $x^{6}$ to $x^{2}$ because $(x^{6})^{1/3} = x^{6*1/3} = x^{2}$ . $\newline$ $\lim_{x \to \infty}\left(\frac{x^{2} \cdot \sqrt[3]{-\frac{41}{x}+1}}{5x(1+x)}\right)$
Divide by $x^{2}$ : Now we divide every term by $x^{2}$ to simplify the limit further. $\lim_{x \to \infty}\left(\frac{\sqrt[3]{-\frac{41}{x}+1}}{\frac{5}{x}+5}\right)$
Evaluate the limit: As $x$ approaches infinity, $-\frac{41}{x}$ approaches $0$ and $\frac{5}{x}$ approaches $0$ . $\lim_{x \to \infty}\left(\frac{\sqrt[3]{0+1}}{0+5}\right)$
Evaluate the limit: As $x$ approaches infinity, $-\frac{41}{x}$ approaches $0$ and $\frac{5}{x}$ approaches $0$ .
$\lim_{x \to \infty}(\sqrt[3]{0+1})/(0+5)$ Now we can evaluate the limit:
$\lim_{x \to \infty}(\sqrt[3]{1})/5 = \frac{1}{5}$