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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)(64x^(10)+2x^(6)+41x^(4)))/(3x^(4)+3x^(3)+8x^(2))
Answer:

Determine the following limit in simplest form. If the limit is infinite, state that the limit does not exist (DNE). $\newline$ $\lim _{x \rightarrow \infty} \frac{\sqrt[3]{64 x^{10}+2 x^{6}+41 x^{4}}}{3 x^{4}+3 x^{3}+8 x^{2}}$ $\newline$ 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]{64 x^{10}+2 x^{6}+41 x^{4}}}{3 x^{4}+3 x^{3}+8 x^{2}}

\newline

Answer:

Divide by $x^4$ : We are given the limit: $\lim_{x \to \infty}\left(\frac{\sqrt[3]{64x^{10}+2x^{6}+41x^{4}}}{3x^{4}+3x^{3}+8x^{2}}\right)$ To simplify this limit, we will first divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^4$ .
Simplify terms: Divide each term in the numerator and the denominator by $x^4$ : $\lim_{x \to \infty}\left(\frac{\sqrt[3]{\frac{64x^{10}}{x^4} + \frac{2x^{6}}{x^4} + \frac{41x^{4}}{x^4}}}{\frac{3x^{4}}{x^4} + \frac{3x^{3}}{x^4} + \frac{8x^{2}}{x^4}}\right)$
Ignore smaller terms: Simplify each term: $\lim_{x \to \infty}\left(\frac{\sqrt[3]{64x^{6} + 2x^{2} + 41}}{3 + \frac{3}{x} + \frac{8}{x^2}}\right)$
Take cube root: As $x$ approaches infinity, the terms $\frac{2x^{2}}{x^4}$ , $\frac{41}{x^4}$ , $\frac{3}{x}$ , and $\frac{8}{x^2}$ will approach zero. Therefore, we can ignore these terms for the limit calculation: $\newline$ $\lim_{x \to \infty}\left(\frac{\sqrt[3]{64x^{6}}}{3}\right)$
Calculate limit: Now, take the cube root of $64x^{6}$ : $\newline$ $\lim_{x \to \infty}\left(\frac{4x^{2}}{3}\right)$
Limit does not exist: Since $x^{2}$ grows without bound as $x$ approaches infinity, the limit of rac{4x^{2}}{3} as $x$ approaches infinity is also infinity. Therefore, the limit does not exist (DNE).

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