Compute $\displaystyle \lim _{x\to -\infty }{\frac {3x^4-4x^3-2x+1}{8x^4+9x^2+5}}$. (Use I to stand for $\infty $ if needed.)

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We start by restating the problem: compute the limit as x approaches negative infinity of (3x^4 - 4x^3 - 2x + 1) / (8x^4 + 9x^2 + 5).
A common strategy for rational expressions with polynomials is to compare the highest-degree terms, since they dominate the behavior as x grows without bound in magnitude.
Option A for a quick route is to divide every term by x^4, the highest power present in both numerator and denominator. This yields:
(3 - 4/x - 2/x^3 + 1/x^4) / (8 + 9/x^2 + 5/x^4).
As x → -∞, the fractions involving 1/x, 1/x^2, 1/x^3, and 1/x^4 all tend to zero, because the denominators grow without bound while the numerators stay finite.
Consequently, the expression simplifies in the limit to (3 - 0 - 0 + 0) / (8 + 0 + 0) = 3/8, given that x^4 is positive for large magnitudes of x, so the signs induced by odd powers in the lower-degree terms vanish in the leading behavior.
Alternatively, one can reason by the standard leading-term comparison: the highest-degree terms in numerator and denominator are 3x^4 and 8x^4, and for large |x|, lower-degree terms become negligible in comparison; thus the ratio tends toward 3/8.
Note that the direction of x approaching -∞ does not affect x^4, since x^4 is an even power and remains positive, so the limit from the left and right agree in this case.
If needed, this can also be framed as a limit of a simplified function obtained by factoring out x^4: (3 - 4/x - 2/x^3 + 1/x^4) / (8 + 9/x^2 + 5/x^4) and evaluating the limit term-by-term, reinforcing that all terms with 1/x^k vanish as x → -∞.
In summary, the dominant contribution comes from the leading coefficients 3 in the numerator and 8 in the denominator, with all lower-degree terms vanishing in the limit.

_MATH1013_1ABCD_2025 Subsection 3.2 (closed on 4 Oct)

Compute \(\displaystyle \lim _{x\to -\infty }{\frac {3x^4-4x^3-2x+1}{8x^4+9x^2+5}}\). (Use I to stand for \(\infty \) if needed.)

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