Math Problem Statement

Find \( \lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} \).

Solution

To evaluate the limit

limxlnx2x\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}}

we'll analyze the behavior of the function as xx approaches infinity.

Step 1: Apply L'Hôpital's Rule

Since both lnx\ln x and 2x2\sqrt{x} grow without bounds as xx \to \infty, we have an indeterminate form of \frac{\infty}{\infty}, which allows us to apply L'Hôpital's Rule. This rule states that for functions of the form f(x)g(x)\frac{f(x)}{g(x)} where both f(x)f(x) \to \infty and g(x)g(x) \to \infty as xx \to \infty, we can find the limit by differentiating the numerator and the denominator.

Step 2: Differentiate the Numerator and the Denominator

  1. Differentiate the numerator, lnx\ln x: ddx(lnx)=1x\frac{d}{dx}(\ln x) = \frac{1}{x}

  2. Differentiate the denominator, 2x2\sqrt{x}: ddx(2x)=ddx(2x1/2)=212x1/2=1x\frac{d}{dx}(2\sqrt{x}) = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{-1/2} = \frac{1}{\sqrt{x}}

Step 3: Rewrite the Limit Using the Derivatives

The limit now becomes: limx1x1x=limx1xx=limxxx\lim_{{x \to \infty}} \frac{\frac{1}{x}}{\frac{1}{\sqrt{x}}} = \lim_{{x \to \infty}} \frac{1}{x} \cdot \sqrt{x} = \lim_{{x \to \infty}} \frac{\sqrt{x}}{x}

Step 4: Simplify the Expression

Rewrite xx\frac{\sqrt{x}}{x} as: xx=x1/2x=x1/2=1x\frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x} = x^{-1/2} = \frac{1}{\sqrt{x}}

Step 5: Take the Limit

Now, as xx \to \infty, 1x0\frac{1}{\sqrt{x}} \to 0.

Therefore: limxlnx2x=0\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} = 0

Final Answer

limxlnx2x=0\lim_{{x \to \infty}} \frac{\ln x}{2\sqrt{x}} = 0

Would you like more details or have any questions?

Here are five related questions for further exploration:

  1. What is L'Hôpital's Rule and when can it be applied?
  2. How does the growth rate of lnx\ln x compare to xαx^{\alpha} for various values of α\alpha?
  3. Can we generalize this limit if the denominator is changed to xpx^p for some p>0p > 0?
  4. What are other methods to analyze limits of the form lnxxp\frac{\ln x}{x^p} as xx \to \infty?
  5. How does the behavior of logarithmic functions compare to polynomial and exponential functions as xx approaches infinity?

Tip: For limits involving lnx\ln x and powers of xx, it can often help to compare their growth rates, as polynomial functions tend to grow faster than logarithmic functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Logarithmic and Radical Functions

Formulas

L'Hôpital's Rule
Differentiation rules for \( \ln x \) and \( x^{1/2} \)

Theorems

L'Hôpital's Rule

Suitable Grade Level

Undergraduate Calculus