Compute [math: limx→−∞x3e−3x]\displaystyle \lim _{x o -\infty }{\frac{x^3}{e^{-3x}}}. (Use I to stand for [math: ∞]\infty if needed.)

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Compute [math: limx→−∞x3e−3x]\displaystyle \lim _{x	o -\infty }{\frac{x^3}{e^{-3x}}}. (Use I to stand for [math: ∞]\infty if needed.)

BlackTom AI · Accepted Answer

Let me restate the problem in my own words: we want the limit as x approaches negative infinity of x^3 divided by e^{−3x}, i.e., lim_{x→−∞} (x^3 / e^{−3x}).

First, notice that e^{−3x} can be rewritten as e^{3|x|}, which grows very large as x → −∞, while x^3 grows in magnitude only polynomially (to −∞). So we’re dealing with a quotient of a polynomial tending to −∞ and an exponential term tending to ∞. This is a classic situation where the exponential dominates the polynomial in terms of growth/decay, but we should handle it carefully since the signs are involved.

A standard approach is to use L'Hôpital's rule by rewriting the expression in a form suitable for it. Observe: x^3 / e^{−3x} is an ∞/∞ type as x → −∞ (numerator → −∞, denominator → ∞). Differentiate numerator and denominator:
- Numerator derivative: d(x^3)/dx = 3x^2
- Denominator derivative: d(e^{−3x})/dx = −3 e^{−3x}
Thus the limit becomes lim_{x→−∞} [3x^2 / (−3 e^{−3x})] = lim_{x→−∞} [−x^2 / e^{−3x}].

Now, the new form still has the type ∞/∞ (since x^2 → ∞ and e^{−3x} → ∞ as x → −∞). We can apply L'Hôpital's rule again:
- Numerator derivative: d(−x^2)/dx = −2x
- Denominator derivative: d(e^{−3x})/dx = −3 e^{−3x}
So it becomes lim_{x→−∞} [−2x / (−3 e^{−3x})] = lim_{x→−∞} [ (2x) / (3 e^{−3x}) ].

A third application may simplify further, but an alternative route is to rewrite the expression in a form that isolates the exponential decay. Notice that e^{−3x} = e^{3x}⁻¹, so x^3 / e^{−3x} = x^3 e^{3x}. This representation is often easier to analyze as x → −∞ because e^{3x} → 0 and x^3 → −∞, giving a product of a polynomial going to −∞ and an exponential going to 0. In this product form, the exponential decay to 0 dominates the polynomial growth, driving the product to 0.

To make the domination argument precise, consider the magnitude |x^3 e^{3x}| = |x^3| e^{3x}. Since e^{3x} decays faster than any power of |x| grows as x → −∞, the product |x^3| e^{3x} → 0. Therefore x^3 e^{3x} → 0, and hence x^3 / e^{−3x} → 0. The sign follows from x^3 being negative for large negative x and e^{−3x} being positive, but the limit of the whole expression is 0 regardless of sign.

Consequently, the limit evaluates to 0.

_MATH1013_1ABCD_2025 Subsection 4.3 (closed on 1 Nov)

Compute [math: limx→−∞x3e−3x]\displaystyle \lim _{x\to -\infty }{\frac{x^3}{e^{-3x}}}. (Use I to stand for [math: ∞]\infty if needed.)

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