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\( \begin{array}{lll}\text { (a) } \lim _{x \rightarrow-\infty} \frac{3 x^{6}-7 x^{5}+x}{5 x^{6}+4 x^{5}-3} & \text { (b) } \lim _{x \rightarrow \infty} \frac{5 x+2 x^{3}}{x^{3}+x-7} & \text { (c) } \lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^{2}+3 x}-7}{7-3 x} \\ \text { (d) } \lim _{x \rightarrow \infty} \frac{x^{2}}{e^{4 x}-1-4 x} & \text { (e) } \lim _{x \rightarrow \infty} \frac{1+3 x}{\sqrt{2 x^{2}+x}} & \text { (f) } \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-\sqrt{x^{2}+1}\right)\end{array} \)

Ask by Kelly Valdez. in South Africa
Mar 12,2025

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(a) \( \frac{3}{5} \) (b) \( 2 \) (c) \( \frac{2}{3} \) (d) \( 0 \) (e) \( \frac{3}{\sqrt{2}} \) (f) \( -\frac{1}{2} \)

Solution

(a) We have \[ \lim_{x\rightarrow-\infty}\frac{3x^6-7x^5+x}{5x^6+4x^5-3}. \] Factor out \(x^6\) in both numerator and denominator: \[ \frac{x^6\left(3-7/x+1/x^5\right)}{x^6\left(5+4/x-3/x^6\right)} = \frac{3-7/x+1/x^5}{5+4/x-3/x^6}. \] Taking the limit as \(x\rightarrow-\infty\) (where the terms with \(1/x\) vanish), we obtain \[ \frac{3}{5}. \] --- (b) We have \[ \lim_{x\rightarrow \infty}\frac{5x+2x^3}{x^3+x-7}. \] Divide numerator and denominator by \(x^3\): \[ \frac{\frac{5}{x^2}+2}{1+\frac{1}{x^2}-\frac{7}{x^3}}. \] As \(x\rightarrow\infty\), the terms with \(1/x^2\) and \(1/x^3\) vanish, so the limit becomes \[ \frac{2}{1}=2. \] --- (c) We have \[ \lim_{x\rightarrow-\infty}\frac{\sqrt{4x^2+3x}-7}{7-3x}. \] First, factor inside the square root: \[ \sqrt{4x^2+3x} = \sqrt{4x^2\left(1+\frac{3}{4x}\right)} = 2|x|\sqrt{1+\frac{3}{4x}}. \] For \(x\rightarrow -\infty\), we have \(|x|=-x\), so \[ \sqrt{4x^2+3x} = -2x\sqrt{1+\frac{3}{4x}}. \] Thus, the expression becomes \[ \frac{-2x\sqrt{1+\frac{3}{4x}}-7}{7-3x}. \] For very large negative \(x\), the dominant term in the numerator is \(-2x\sqrt{1+\frac{3}{4x}} \approx -2x\) and in the denominator it is \(-3x\). Therefore, \[ \lim_{x\rightarrow-\infty} \frac{-2x}{-3x}=\frac{2}{3}. \] --- (d) We have \[ \lim_{x\rightarrow\infty}\frac{x^2}{e^{4x}-1-4x}. \] Since \(e^{4x}\) grows exponentially while \(x^2\) grows polynomially, the denominator becomes infinitely large compared to the numerator. Hence, the limit is \[ 0. \] --- (e) We have \[ \lim_{x\rightarrow\infty}\frac{1+3x}{\sqrt{2x^2+x}}. \] Divide numerator and denominator by \(x\). In the numerator, factor out \(x\) (for large \(x\), \(1\) is negligible): \[ 1+3x \sim 3x. \] In the denominator, write \[ \sqrt{2x^2+x} = x\sqrt{2+\frac{1}{x}} \sim x\sqrt{2}. \] Thus, \[ \lim_{x\rightarrow\infty} \frac{3x}{x\sqrt{2}}=\frac{3}{\sqrt{2}}. \] --- (f) We have \[ \lim_{x\rightarrow\infty}\left(\sqrt{x^2-x+1}-\sqrt{x^2+1}\right). \] Rationalize the expression by multiplying and dividing by the conjugate: \[ \sqrt{x^2-x+1}-\sqrt{x^2+1} = \frac{(\sqrt{x^2-x+1}-\sqrt{x^2+1})(\sqrt{x^2-x+1}+\sqrt{x^2+1})}{\sqrt{x^2-x+1}+\sqrt{x^2+1}}. \] The numerator becomes \[ (x^2-x+1) - (x^2+1) = -x. \] Thus, the expression is \[ \frac{-x}{\sqrt{x^2-x+1}+\sqrt{x^2+1}}. \] For large \(x\), \[ \sqrt{x^2-x+1} \sim x \quad \text{and} \quad \sqrt{x^2+1} \sim x, \] so the denominator is approximately \(2x\). Therefore, \[ \lim_{x\rightarrow\infty}\frac{-x}{2x}=-\frac{1}{2}. \]

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Bonus Knowledge

To tackle the given limits, it's helpful to identify the leading terms since they dominate the behavior of the functions as \( x \) approaches infinity or negative infinity. For part (a), notice that the leading term in both the numerator and denominator is \( 3x^6 \) and \( 5x^6 \) respectively, yielding a limit of \( \frac{3}{5} \) as \( x \to -\infty \). For part (b), as \( x \) approaches infinity, the leading term in the numerator is \( 2x^3 \) and in the denominator is also \( x^3 \). Thus, the limit simplifies to \( \frac{2}{1} = 2 \). Keep in mind that for functions with square roots and polynomial expressions, isolating the dominant terms is key, especially when analyzing limits that trend towards infinity or negative infinity! For further challenges, re-examine limits where there are both rational and irrational components—strategically breaking them down by rationalizing where necessary can yield clearer insights into their behavior as they approach critical points. Cozying up to those tricky square roots in limits doesn’t just help in calculations but can also reveal elegant relationships within the function!

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