Which statement best describes the limit shown below? lim_(x rarr oo)(2x^(16)-3x^(24))/(-2x^(24)+log_(5)x) The limit equals zero The limit does not exist The limit exists and does not equal zero

Analyze Behavior as x Approaches Infinity: We need to analyze the behavior of the function as x approaches infinity. To do this, we will compare the degrees of the polynomials in the numerator and the denominator.Compare Highest Degree Terms: The highest degree term in the numerator is -3x^(24) and in the denominator is -2x^(24). As x approaches infinity, the terms with lower degrees become insignificant compared to the highest degree terms.Simplify by Dividing Highest Power: We can simplify the limit by dividing both the numerator and the denominator by x^(24), the highest power of x present in the expression. lim_(x rarr oo)(2x^(16)-3x^(24))/(-2x^(24)+log_(5)x) = lim_(x rarr oo)(2/x^(8)-3)/(-2+log_(5)x/x^(24))Simplify as x Approaches Infinity: As x approaches infinity, 2/x^(8) and log_(5)x/x^(24) both approach 0. Therefore, the limit simplifies to: lim_(x rarr oo)(-3)/(-2) = 3/2Conclusion: The limit exists and does not equal zero. The correct statement is ＂The limit exists and does not equal zero.＂

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Which statement best describes the limit shown below?

lim_(x rarr oo)(2x^(16)-3x^(24))/(-2x^(24)+log_(5)x)
The limit equals zero
The limit does not exist
The limit exists and does not equal zero

Which statement best describes the limit shown below? $\newline$ $\lim _{x \rightarrow \infty} \frac{2 x^{16}-3 x^{24}}{-2 x^{24}+\log _{5} x}$ $\newline$ The limit equals zero $\newline$ The limit does not exist $\newline$ The limit exists and does not equal zero

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Q. Which statement best describes the limit shown below?

\newline

\lim _{x \rightarrow \infty} \frac{2 x^{16}-3 x^{24}}{-2 x^{24}+\log _{5} x}

\newline

The limit equals zero

\newline

The limit does not exist

\newline

The limit exists and does not equal zero

Analyze Behavior as $x$ Approaches Infinity: We need to analyze the behavior of the function as $x$ approaches infinity. To do this, we will compare the degrees of the polynomials in the numerator and the denominator.
Compare Highest Degree Terms: The highest degree term in the numerator is $-3x^{24}$ and in the denominator is $-2x^{24}$ . As $x$ approaches infinity, the terms with lower degrees become insignificant compared to the highest degree terms.
Simplify by Dividing Highest Power: We can simplify the limit by dividing both the numerator and the denominator by $x^{24}$ , the highest power of $x$ present in the expression. $\newline$ $\lim_{x \to \infty}\frac{2x^{16}-3x^{24}}{-2x^{24}+\log_{5}x} = \lim_{x \to \infty}\frac{\frac{2}{x^{8}}-3}{-2+\frac{\log_{5}x}{x^{24}}}$
Simplify as $x$ Approaches Infinity: As $x$ approaches infinity, $\frac{2}{x^{8}}$ and $\frac{\log_{5}x}{x^{24}}$ both approach $0$ . Therefore, the limit simplifies to: $\lim_{x \rightarrow \infty}\left(\frac{-3}{-2}\right) = \frac{3}{2}$
Conclusion: The limit exists and does not equal zero. The correct statement is ＂The limit exists and does not equal $0$ .＂