Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE). $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt{-11 x^{2}+22 x+4 x^{4}}}{9 x+4 x^{3}}$ $\newline$ 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{-11 x^{2}+22 x+4 x^{4}}}{9 x+4 x^{3}}

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Answer:

Factor Out Highest Powers: We need to find the limit of the given function as $x$ approaches infinity. The function is $\frac{\sqrt{-11x^{2}+22x+4x^{4}}}{9x+4x^{3}}$ . To simplify the limit, we will factor out the highest power of $x$ in the numerator and the denominator.
Simplify Numerator and Denominator: In the numerator, the highest power of $x$ is $x^4$ under the square root, which is equivalent to $x^2$ when taken outside the square root. In the denominator, the highest power of $x$ is $x^3$ . We factor these out from the numerator and denominator respectively.
Cancel Out $x^2$ : After factoring out the highest powers of $x$ , we have: $\newline$ $\lim_{x \to \infty}\left(\frac{x^2 \sqrt{4 - \frac{11}{x^2} + \frac{22}{x^3}}}{x^3 \left(4 + \frac{9}{x^2}\right)}\right)$
Simplify Further: We can now simplify the expression by canceling out an $x^2$ from the numerator and denominator: $\lim_{x \rightarrow \infty}\left(\frac{\sqrt{4 - \frac{11}{x^2} + \frac{22}{x^3}}}{x \cdot (4 + \frac{9}{x^2})}\right)$
Final Limit is $0$ : As $x$ approaches infinity, the terms with $x$ in the denominator ( $\frac{11}{x^2}$ , $\frac{22}{x^3}$ , and $\frac{9}{x^2}$ ) approach $0$ . We can then simplify the limit to: $\newline$ $\lim_{x \rightarrow \infty}(\frac{\sqrt{4}}{x \cdot (4)})$
Final Limit is $0$ : As $x$ approaches infinity, the terms with $x$ in the denominator ( $\frac{11}{x^2}$ , $\frac{22}{x^3}$ , and $\frac{9}{x^2}$ ) approach $0$ . We can then simplify the limit to: $\newline$ $\lim_{x \rightarrow \infty}\left(\frac{\sqrt{4}}{x \cdot (4)}\right)$ The square root of $4$ is $2$ , so the limit simplifies further to: $\newline$ $\lim_{x \rightarrow \infty}\left(\frac{2}{4x}\right)$
Final Limit is $0$ : As $x$ approaches infinity, the terms with $x$ in the denominator ( $\frac{11}{x^2}$ , $\frac{22}{x^3}$ , and $\frac{9}{x^2}$ ) approach $0$ . We can then simplify the limit to: $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt{4}}{x \cdot (4)}\right)$ The square root of $4$ is $2$ , so the limit simplifies further to: $\newline$ $\lim_{x \to \infty}\left(\frac{2}{4x}\right)$ As $x$ approaches infinity, $\frac{2}{4x}$ approaches $0$ . Therefore, the limit of the original function as $x$ approaches infinity is $0$ .