Compute [math: limx→−∞3x4−4x3−2x+18x4+9x2+5]\displaystyle \lim _{x o -\infty }{\frac {3x^4-4x^3-2x+1}{8x^4+9x^2+5}}. (Use I to stand for [math: ∞]\infty if needed.)

Question

Compute [math: limx→−∞3x4−4x3−2x+18x4+9x2+5]\displaystyle \lim _{x	o -\infty }{\frac {3x^4-4x^3-2x+1}{8x^4+9x^2+5}}. (Use I to stand for [math: ∞]\infty if needed.)

BlackTom AI · Accepted Answer

We start by examining the given limit: lim_{x→-∞} (3x^4 - 4x^3 - 2x + 1) / (8x^4 + 9x^2 + 5).
A standard trick for rational functions as x grows without bound is to divide every term by the highest power of x present in the denominator, here x^4.
Dividing numerator and denominator by x^4 gives:
(3 - 4/x - 2/x^3 + 1/x^4) / (8 + 9/x^2 + 5/x^4).
Now, as x → -∞, each term with a factor 1/x^k tends to 0 (since powers of x in the denominator grow without bound). Therefore the expression simplifies to (3 - 0 - 0 + 0) / (8 + 0 + 0) = 3/8.
A useful check is to note that x^4 is positive for large magnitude x regardless of the sign of x, so the leading terms 3x^4 in the numerator and 8x^4 in the denominator dominate and their ratio approaches 3/8. Any lower-degree terms vanish in the limit, confirming the result.
In summary, after normalization by x^4, all vanishing terms disappear, leaving the constant ratio 3/8 as x tends to negative infinity.

Compute [math: limx→−∞3x4−4x3−2x+18x4+9x2+5]\displaystyle \lim _{x\to -\infty }{\frac {3x^4-4x^3-2x+1}{8x^4+9x^2+5}}. (Use I to stand for [math: ∞]\infty if needed.)简答题

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