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5. Explain how to sind the value for each limit a) \( \lim _{x \rightarrow-\infty}-\frac{2 x^{4}+8 x^{5}+7}{3 x^{5}+3 x^{3}-1} \)

Ask by Cross Weston. in the United States
Feb 26,2025

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Answer

The limit is –8/3.

Solution

To evaluate   lim₍ₓ→–∞₎ –[(2x⁴ + 8x⁵ + 7) / (3x⁵ + 3x³ – 1)], follow these steps: 1. Identify the highest power of x in the numerator and the denominator.   • In the numerator, the highest power is x⁵ (the term 8x⁵).   • In the denominator, the highest power is also x⁵ (the term 3x⁵). 2. Factor x⁵ from both the numerator and the denominator.   Rewrite the numerator:    2x⁴ + 8x⁵ + 7 = x⁵(2/x + 8 + 7/x⁵).   Rewrite the denominator:    3x⁵ + 3x³ – 1 = x⁵(3 + 3/x² – 1/x⁵). 3. Simplify the expression by canceling the common factor x⁵ (valid as x ≠ 0):   The expression becomes    – [ (2/x + 8 + 7/x⁵) / (3 + 3/x² – 1/x⁵) ]. 4. Take the limit as x → –∞.   As x → –∞, the terms with 1/x, 1/x², and 1/x⁵ will approach 0.   Thus, the expression inside the bracket simplifies to    (0 + 8 + 0) / (3 + 0 – 0) = 8/3. 5. Don’t forget the negative sign out front.   The final value of the limit is    –(8/3) = –8/3. Thus,   lim₍ₓ→–∞₎ –[(2x⁴ + 8x⁵ + 7) / (3x⁵ + 3x³ – 1)] = –8/3.

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To find the limit as \( x \) approaches \(-\infty\), we should identify the leading terms in both the numerator and denominator. In this case, the numerator \( -\frac{2x^4 + 8x^5 + 7}{3x^5 + 3x^3 - 1} \) can be simplified by dividing every term by \( x^5 \) (the highest power in the denominator). This gives us: \[ \lim_{x \rightarrow -\infty} -\frac{\frac{2}{x} + 8 + \frac{7}{x^5}}{3 + \frac{3}{x^2} - \frac{1}{x^5}} \] As \( x \) approaches \(-\infty\), the terms \(\frac{2}{x}\), \(\frac{7}{x^5}\), \(\frac{3}{x^2}\), and \(-\frac{1}{x^5}\) all approach zero. Therefore, the limit simplifies to: \[ -\frac{8}{3} \] So, the final answer is: \[ \lim _{x \rightarrow-\infty}-\frac{2 x^{4}+8 x^{5}+7}{3 x^{5}+3 x^{3}-1} = -\frac{8}{3} \] Now, you’ve got a handle on limits at infinity! If you need to tackle more tricky limits, consider analyzing behavior at extreme values and simplifying expressions—who knew math could be a smooth ride?

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