Compute [math: limx→−∞x4−8x3−3x+56x4+7x2+3]\displaystyle \lim _{x o -\infty }{\frac {x^4-8x^3-3x+5}{6x^4+7x^2+3}}. (Use I to stand for [math: ∞]\infty if needed.)

Question

Compute [math: limx→−∞x4−8x3−3x+56x4+7x2+3]\displaystyle \lim _{x	o -\infty }{\frac {x^4-8x^3-3x+5}{6x^4+7x^2+3}}. (Use I to stand for [math: ∞]\infty if needed.)

BlackTom AI · Accepted Answer

As x approaches negative infinity, the highest power term in both numerator and denominator will dominate the expression, so we compare the leading x^4 terms.

Optionally factor out x^4 from numerator and denominator to simplify the limit:

(x^4 - 8x^3 - 3x + 5) / (6x^4 + 7x^2 + 3) = [x^4(1 - 8/x - 3/x^3 + 5/x^4)] / [x^4(6 + 7/x^2 + 3/x^4)].

Cancel the common x^4 factor:

(1 - 8/x - 3/x^3 + 5/x^4) / (6 + 7/x^2 + 3/x^4).

Now evaluate the limit as x → -∞. Terms with 1/x, 1/x^2, 1/x^3, and 1/x^4 all tend to 0, so the expression tends to (1 - 0 - 0 + 0) / (6 + 0 + 0) = 1/6.

Key points to avoid mistakes:
- Do not confuse the sign of x when only even powers remain; x^4 is always positive, so the leading terms dictate the sign and magnitude regardless of approaching from the left or right.
- If one forgets to factor out x^4 or mistakenly keeps lower-order terms, the limit calculation may become messy or incorrect.
- Failing to recognize that terms like -8/x vanish in the limit would lead to an incorrect final value; only the leading x^4 terms survive in the ratio.

Thus, the limit is determined by the ratio of the leading coefficients of the highest power terms, namely 1 (numerator) over 6 (denominator).

_MATH1013_1ABCD_2025 Subsection 3.2 (closed on 4 Oct)

Compute [math: limx→−∞x4−8x3−3x+56x4+7x2+3]\displaystyle \lim _{x\to -\infty }{\frac {x^4-8x^3-3x+5}{6x^4+7x^2+3}}. (Use I to stand for [math: ∞]\infty if needed.)

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