Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE). $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt{9 x^{6}+20 x^{4}}}{10 x^{3}+6 x}$ $\newline$ 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{9 x^{6}+20 x^{4}}}{10 x^{3}+6 x}

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Answer:

Factor out highest power of x: We are given the limit expression $\lim_{x \to \infty}(\sqrt{9x^{6}+20x^{4}})/(10x^{3}+6x)$ . To simplify this, we will first factor out the highest power of $x$ in the numerator and denominator.
Rewrite with factored terms: In the numerator, the highest power of $x$ inside the square root is $x^6$ . We factor out $x^6$ from the square root, which is equivalent to $x^3$ outside the square root. $\newline$ $\sqrt{9x^{6}+20x^{4}} = \sqrt{x^{6}(9+20/x^{2})} = x^3 \cdot \sqrt{9+20/x^{2}}$
Cancel out $x^3$ terms: In the denominator, the highest power of $x$ is $x^3$ . We factor out $x^3$ from the entire denominator. $\frac{10x^{3}+6x}{x^3(10+\frac{6}{x^2})}$
Approach infinity: Now we rewrite the limit expression with the factored terms. $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt{9x^{6}+20x^{4}}}{10x^{3}+6x}\right) = \lim_{x \to \infty}\left(\frac{x^3 \sqrt{9+20/x^{2}}}{x^3(10+6/x^2)}\right)$
Simplify square root: We can now cancel out the $x^3$ terms in the numerator and the denominator.\lim_{x \to \infty}\left(\frac{x^$3$ \sqrt{$9$+\frac{$20$}{x^{$2$}}}}{x^$3$($10$+\frac{$6$}{x^$2$})}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{$9$+\frac{$20$}{x^{$2$}}}}{$10$+\frac{$6$}{x^$2$}}\right)
Simplify square root: We can now cancel out the $x^3$ terms in the numerator and the denominator.$\lim_{x \to \infty}\left(\frac{x^3 \sqrt{9+\frac{20}{x^{2}}}}{x^3(10+\frac{6}{x^2})}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{9+\frac{20}{x^{2}}}}{10+\frac{6}{x^2}}\right)$As $x$ approaches infinity, the terms $\frac{20}{x^2}$ and $\frac{6}{x^2}$ in the limit expression will approach zero.$\lim_{x \to \infty}\left(\frac{\sqrt{9+\frac{20}{x^{2}}}}{10+\frac{6}{x^2}}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{9+0}}{10+0}\right) = \frac{\sqrt{9}}{10}$
Simplify square root: We can now cancel out the $x^3$ terms in the numerator and the denominator.$\newline$\[\lim_{x \to \infty}\left(\frac{x^3 \sqrt{9+\frac{20}{x^{2}}}}{x^3(10+\frac{6}{x^2})}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{9+\frac{20}{x^{2}}}}{10+\frac{6}{x^2}}\right)As $x$ approaches infinity, the terms $\frac{20}{x^2}$ and $\frac{6}{x^2}$ in the limit expression will approach zero. $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt{9+\frac{20}{x^{2}}}}{10+\frac{6}{x^2}}\right) = \lim_{x \to \infty}\left(\frac{\sqrt{9+0}}{10+0}\right) = \frac{\sqrt{9}}{10}$ Simplify the square root of $9$ , which is $3$ . $\newline$ $\frac{\sqrt{9}}{10} = \frac{3}{10}$