Compute $\displaystyle \lim _{x\to -\infty }{\frac {x^4-8x^3-3x+5}{6x^4+7x^2+3}}$. (Use I to stand for $\infty $ if needed.)

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Question: Compute the limit as x approaches negative infinity of (x^4 - 8x^3 - 3x + 5) / (6x^4 + 7x^2 + 3).

Step 1: Identify the dominant terms for large |x|. Both numerator and denominator are polynomials with leading term x^4, so the behavior is governed by these highest-degree terms.

Step 2: Divide every term by x^4 to normalize. The expression becomes:
   (1 - 8/x - 3/x^3 + 5/x^4) / (6 + 7/x^2 + 3/x^4).

Step 3: Evaluate the limit as x → -∞. Each of the terms with x in the denominator tends to 0:
   -8/x → 0, -3/x^3 → 0, 5/x^4 → 0, 7/x^2 → 0, 3/x^4 → 0.

Step 4: The expression simplifies to (1 - 0 - 0 + 0) / (6 + 0 + 0) = 1/6.

Notes: The sign of x does not affect the leading term because x^4 is always nonnegative for large |x|, so the limit is determined entirely by the ratio of the leading coefficients 1 (from x^4 in the numerator) and 6 (from x^4 in the denominator).

_MATH1013_1ABCD_2025 Subsection 3.2 (closed on 4 Oct)

Compute \(\displaystyle \lim _{x\to -\infty }{\frac {x^4-8x^3-3x+5}{6x^4+7x^2+3}}\). (Use I to stand for \(\infty \) if needed.)

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