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

lim_(x rarr oo)(root(3)(-33 x-27x^(5)-33))/(5+9x+5x^(2))
Answer:

Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE). $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-33 x-27 x^{5}-33}}{5+9 x+5 x^{2}}$ $\newline$ Answer:

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Full solution

Q. 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]{-33 x-27 x^{5}-33}}{5+9 x+5 x^{2}}

\newline

Answer:

Identify highest power: We have the limit: $\lim_{x \to \infty}\left(\sqrt[3]{-33x - 27x^5 - 33}\right)/(5 + 9x + 5x^2)$ First, we observe the highest power of $x$ in the numerator and the denominator to simplify the expression.
Divide by $x^2$ : In the numerator, the highest power of $x$ is $x^5$ , and in the denominator, the highest power of $x$ is $x^2$ . To simplify, we divide both the numerator and the denominator by $x^2$ .
Rewrite the limit: We rewrite the limit as: $\lim_{x \to \infty}\left(\frac{\sqrt[3]{-\frac{33}{x^4} - 27x^3 - \frac{33}{x^7}}}{\frac{5}{x^2} + \frac{9}{x} + 5}\right)$
Ignore terms with $x$ : As $x$ approaches infinity, the terms with $x$ in the denominator will approach zero. So, we can ignore the terms $-\frac{33}{x^4}$ and $-\frac{33}{x^7}$ in the numerator and $\frac{5}{x^2}$ and $\frac{9}{x}$ in the denominator.
Simplify the limit: The limit simplifies to: \lim_{x \to \infty}\left(\sqrt[\(3]{ $-27$ x^ $3$ }\right)/ $5$
Take cube root: We can now take the cube root of $-27x^3$ , which simplifies to $-3x$ .
Final limit: The limit is now: $\lim_{x \to \infty} \frac{-3x}{5}$
Approach negative infinity: As $x$ approaches infinity, $-\frac{3x}{5}$ will also approach negative infinity.
Limit does not exist: Therefore, the limit does not exist (DNE) because the function approaches negative infinity.

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