Compute [math: limx→−∞−4x3+2x2+6x−57x4+4x2+1]\displaystyle \lim _{x o -\infty }{\frac {-4x^3+2x^2+6x-5}{7x^4+4x^2+1}}. (Use I to stand for [math: ∞]\infty if needed.)

Question

Compute [math: limx→−∞−4x3+2x2+6x−57x4+4x2+1]\displaystyle \lim _{x	o -\infty }{\frac {-4x^3+2x^2+6x-5}{7x^4+4x^2+1}}. (Use I to stand for [math: ∞]\infty if needed.)

BlackTom AI · Accepted Answer

To evaluate the limit as x approaches −∞ of (-4x^3 + 2x^2 + 6x − 5) / (7x^4 + 4x^2 + 1), we should compare the leading growth terms in numerator and denominator.

First, note the degrees: the numerator is a cubic (degree 3) and the denominator is quartic (degree 4). As x -> −∞, higher-degree terms dominate, so the behavior of the fraction is governed by the highest powers.

A clean way to see this is to divide every term by x^4, the highest power in the denominator:
- Numerator divided by x^4: (-4x^3)/x^4 + (2x^2)/x^4 + (6x)/x^4 + (-5)/x^4 = (-4/x) + (2/x^2) + (6/x^3) - (5/x^4).
- Denominator divided by x^4: (7x^4)/x^4 + (4x^2)/x^4 + 1/x^4 = 7 + (4/x^2) + (1/x^4).
As x → −∞, all terms with 1/x, 1/x^2, 1/x^3, or 1/x^4 tend to 0.

Therefore, the limit becomes 0 divided by 7, i.e., 0/7 = 0.

This confirms that the fraction tends to 0 as x → −∞, since the denominator’s growth (degree 4) outpaces the numerator’s growth (degree 3).

Compute [math: limx→−∞−4x3+2x2+6x−57x4+4x2+1]\displaystyle \lim _{x\to -\infty }{\frac {-4x^3+2x^2+6x-5}{7x^4+4x^2+1}}. (Use I to stand for [math: ∞]\infty if needed.)简答题

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