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Find the horizontal asympeote \( \lim _{x \rightarrow \infty}=\frac{4 x^{2}-3 x^{2}+6}{\sqrt{x^{4}-7 x^{3}+2}} \)

Ask by Nunez Herrera. in South Africa
Mar 16,2025

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The horizontal asymptote is \( y = 1 \).

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To find the horizontal asymptote of the given function, we can simplify the expression as \( x \) approaches infinity. Start with the original limit: \[ \lim _{x \rightarrow \infty} \frac{4 x^{2}-3 x^{2}+6}{\sqrt{x^{4}-7 x^{3}+2}}. \] First, simplify the numerator: \[ 4x^{2} - 3x^{2} + 6 = x^{2} + 6. \] Now the limit becomes: \[ \lim_{x \rightarrow \infty} \frac{x^{2} + 6}{\sqrt{x^{4} - 7x^{3} + 2}}. \] Next, focus on the denominator. As \( x \) becomes very large, the \( -7x^{3} \) and \( +2 \) become negligible compared to \( x^{4} \): \[ \sqrt{x^{4} - 7x^{3} + 2} \approx \sqrt{x^{4}} = x^{2}. \] Now, we can rewrite the limit as: \[ \lim_{x \rightarrow \infty} \frac{x^{2} + 6}{x^{2}}. \] Dividing the numerator and denominator by \( x^{2} \) gives: \[ \lim_{x \rightarrow \infty} \left(1 + \frac{6}{x^{2}}\right) = 1 + 0 = 1. \] Therefore, the horizontal asymptote is: \[ y = 1. \]

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