Compute limx→−∞x13[math]\displaystyle \lim _{x o -\infty }{x^{\frac {1}{3}}}. (Use I to stand for ∞[math]\infty if needed.)

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Compute limx→−∞x13[math]\displaystyle \lim _{x	o -\infty }{x^{\frac {1}{3}}}. (Use I to stand for ∞[math]\infty if needed.)

BlackTom AI · Accepted Answer

We start by restating the problem to ensure understanding: evaluate the limit as x approaches negative infinity of the cube root of x, i.e., lim_{x→-∞} x^(1/3).

Note about the answer options: the provided data shows there are no multiple-choice options to analyze, only a short answer field with the value -∞ indicated elsewhere. Since there are no options to evaluate, we will instead reason about the limit directly.

Idea: consider the behavior of the cube root function for very large negative x. The cube root function y = x^(1/3) is the inverse of y = y^3, and it preserves the sign of x. For negative x, x^(1/3) is also negative.

Step-by-step reasoning:
- Let x be a large negative number. Then x < 0.
- The cube root reduces the magnitude more slowly than the cube, but it still grows without bound in the negative direction: as |x| → ∞ with x < 0, x^(1/3) → -∞.
- Therefore, the limit of x^(1/3) as x→-∞ is -∞.

Common misconceptions to avoid:
- Thinking the cube root approaches a finite value; this would be true only for cube roots of bounded x, not as x diverges to -∞.
- Confusing with even roots or square roots, which have different end behavior for negative inputs; the cube root is defined for all real x and preserves direction toward -∞ for x→-∞.

Summary of conclusion (conceptual): the function x^(1/3) diverges to negative infinity as x goes to negative infinity, so the limit is -∞.

_MATH1013_1ABCD_2025 Subsection 3.2 (closed on 4 Oct)

Compute limx→−∞x13[math]\displaystyle \lim _{x\to -\infty }{x^{\frac {1}{3}}}. (Use I to stand for ∞[math]\infty if needed.)

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