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

Question

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

BlackTom AI · Accepted Answer

Observation: rewrite the expression to compare rates of growth. The limit is lim_{x->-∞} x^3 / e^{-3x}. Since 1 / e^{-3x} = e^{3x}, this is the same as lim_{x->-∞} x^3 * e^{3x}.

Idea: as x -> -∞, x^3 goes to -∞, but e^{3x} goes to 0 extremely fast (exponential decay). The exponential factor dominates any polynomial, so their product tends to 0.

Alternative view: treat it as a limit of the form (-∞) * 0, where the 0 comes from e^{3x} and the -∞ comes from x^3. Because e^{3x} decays faster than any power of x grows, the product tends to 0 from the negative side (since x^3 is negative for large negative x while e^{3x} is positive). Therefore the limit is 0.

Conclusion (in reasoning terms): the exponential decay wins over polynomial growth in this product, yielding a limit of 0.

_MATH1013_1ABCD_2025 Subsection 4.3 (closed on 1 Nov)

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

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