Math Problem Statement

Compute the following limits: 1) lim(x -> 0+) x ln(x), 2) lim(x -> 0+) (1/x + 1/ln(x))

Solution

Let's break this problem into two parts and compute the limits step by step.

Part 1: lim⁡x→0+(xln⁡x)\lim_{x \to 0^+} (x \ln x)

Analysis:

  • The natural logarithm, ln⁡x\ln x, approaches −∞-\infty as x→0+x \to 0^+, while xx approaches 0+0^+. Therefore, the product xln⁡xx \ln x is of the indeterminate form 0⋅(−∞)0 \cdot (-\infty).
  • Rewrite xln⁡xx \ln x as: xln⁡x=ln⁡x1/x.x \ln x = \frac{\ln x}{1/x}. This transforms the limit into a −∞∞\frac{-\infty}{\infty} form.

Solve using L'Hôpital's Rule:

lim⁡x→0+xln⁡x=lim⁡x→0+ln⁡x1/x.\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}. Differentiate numerator and denominator: Derivative of ln⁡x=1x,Derivative of 1x=−1x2.\text{Derivative of } \ln x = \frac{1}{x}, \quad \text{Derivative of } \frac{1}{x} = -\frac{1}{x^2}. lim⁡x→0+ln⁡x1/x=lim⁡x→0+1x−1x2=lim⁡x→0+−x=0.\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} -x = 0. Thus: lim⁡x→0+xln⁡x=0.\lim_{x \to 0^+} x \ln x = 0.

Part 2: lim⁡x→0+(1x+1ln⁡x)\lim_{x \to 0^+} \left( \frac{1}{x} + \frac{1}{\ln x} \right)

Analysis:

  • For 1x\frac{1}{x}, as x→0+x \to 0^+, 1x→∞\frac{1}{x} \to \infty.
  • For 1ln⁡x\frac{1}{\ln x}, as x→0+x \to 0^+, ln⁡x→−∞\ln x \to -\infty, so 1ln⁡x→0\frac{1}{\ln x} \to 0.
  • Therefore, the dominant term is 1x\frac{1}{x}, and the sum 1x+1ln⁡x\frac{1}{x} + \frac{1}{\ln x} should diverge to ∞\infty.

Formal Calculation:

Separate the terms: lim⁡x→0+(1x+1ln⁡x)=lim⁡x→0+1x+lim⁡x→0+1ln⁡x.\lim_{x \to 0^+} \left( \frac{1}{x} + \frac{1}{\ln x} \right) = \lim_{x \to 0^+} \frac{1}{x} + \lim_{x \to 0^+} \frac{1}{\ln x}.

  1. lim⁡x→0+1x=∞.\lim_{x \to 0^+} \frac{1}{x} = \infty.
  2. lim⁡x→0+1ln⁡x=0.\lim_{x \to 0^+} \frac{1}{\ln x} = 0.

Adding these together: lim⁡x→0+(1x+1ln⁡x)=∞.\lim_{x \to 0^+} \left( \frac{1}{x} + \frac{1}{\ln x} \right) = \infty.

Final Results:

  1. lim⁡x→0+(xln⁡x)=0\lim_{x \to 0^+} (x \ln x) = 0.
  2. lim⁡x→0+(1x+1ln⁡x)=∞\lim_{x \to 0^+} \left( \frac{1}{x} + \frac{1}{\ln x} \right) = \infty.

Would you like a deeper explanation or visualization for these results?

Follow-up Questions:

  1. What are indeterminate forms, and how does L’Hôpital’s rule help resolve them?
  2. Why does ln⁡x→−∞\ln x \to -\infty as x→0+x \to 0^+?
  3. Can xln⁡xx \ln x ever be a positive value as x→0+x \to 0^+?
  4. Why does the term 1x\frac{1}{x} dominate over 1ln⁡x\frac{1}{\ln x} in Part 2?
  5. How would the limits change if x→∞x \to \infty instead of x→0+x \to 0^+?

Tip:

When solving limits with logarithmic or exponential functions, try rewriting terms into a form suitable for L'Hôpital's Rule or dominance analysis for better insight.

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Math Problem Analysis

Mathematical Concepts

Limits
Logarithmic Functions
Indeterminate Forms
L'Hopital's Rule

Formulas

lim(x -> 0+) x ln(x)
lim(x -> 0+) (1/x + 1/ln(x))
L'Hopital's Rule
ln(x) -> -∞ as x -> 0+

Theorems

L'Hopital's Rule

Suitable Grade Level

Grades 11-12

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