Math Problem Statement

lim(x -> 0+) sqrt^3(x) * lnx

Solution

To find the limit of the expression limx0+x3ln(x)\lim_{x \to 0^+} \sqrt[3]{x} \ln(x), we can break down the problem into parts.

We are evaluating:

limx0+x3ln(x)\lim_{x \to 0^+} \sqrt[3]{x} \ln(x)

Step 1: Analyze the behavior of each term as x0+x \to 0^+.

  1. Cube Root of x:
    x3=x1/3\sqrt[3]{x} = x^{1/3}. As x0+x \to 0^+, x1/30x^{1/3} \to 0.

  2. Natural Logarithm of x:
    ln(x)\ln(x) tends to -\infty as x0+x \to 0^+.

Step 2: Combining the two terms.

We have an indeterminate form 0()0 \cdot (-\infty) as x0+x \to 0^+. This suggests we need to analyze it more carefully, possibly using techniques such as L'Hopital's Rule, or rewriting the expression.

Step 3: Apply L'Hopital's Rule.

We can rewrite the expression as:

x3ln(x)=ln(x)x1/3\sqrt[3]{x} \ln(x) = \frac{\ln(x)}{x^{-1/3}}

This is now in the form \frac{-\infty}{\infty}, which is an indeterminate form, and we can apply L'Hopital's Rule. To do that, we differentiate the numerator and denominator with respect to xx:

  • Derivative of the numerator ln(x)\ln(x) is 1x\frac{1}{x}.
  • Derivative of the denominator x1/3x^{-1/3} is 13x4/3-\frac{1}{3} x^{-4/3}.

So, applying L'Hopital's Rule:

limx0+ln(x)x1/3=limx0+1x13x4/3=limx0+3x7/3\lim_{x \to 0^+} \frac{\ln(x)}{x^{-1/3}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{3} x^{-4/3}} = \lim_{x \to 0^+} -3 x^{7/3}

Step 4: Evaluate the limit.

As x0+x \to 0^+, x7/30x^{7/3} \to 0. Therefore, the limit is:

limx0+3x7/3=0\lim_{x \to 0^+} -3 x^{7/3} = 0

Final Answer:

limx0+x3ln(x)=0\lim_{x \to 0^+} \sqrt[3]{x} \ln(x) = 0


Would you like further explanation or have any questions? Here are some related questions:

  1. What happens to limx0+xln(x)\lim_{x \to 0^+} \sqrt{x} \ln(x)?
  2. How does L'Hopital's Rule apply to other indeterminate forms like 000^0 or /\infty/\infty?
  3. How can I apply L'Hopital's Rule to limits involving exponential functions?
  4. Can you explain why ln(x)\ln(x) \to -\infty as x0+x \to 0^+?
  5. How would this limit change if the expression was x3sin(x)\sqrt[3]{x} \sin(x)?

Tip: When you encounter limits of the form 00 \cdot \infty, it's often useful to rewrite them as fractions and apply L'Hopital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
L'Hopital's Rule

Formulas

Cube root: sqrt^3(x) = x^(1/3)
Natural logarithm behavior: ln(x) → -∞ as x → 0+
L'Hopital's Rule: If lim(x → c) f(x)/g(x) = indeterminate form, then lim(x → c) f'(x)/g'(x)

Theorems

L'Hopital's Rule

Suitable Grade Level

Undergraduate Mathematics