Math Problem Statement

lim x->inf ln(x)/x^1/x using lhopital

Solution

To find the limit:

limxln(x)x1/x\lim_{x \to \infty} \frac{\ln(x)}{x^{1/x}}

we can start by simplifying the expression. Let's take it step by step.

Step 1: Simplify the exponent term

Notice that x1/xx^{1/x} is an expression involving an exponent. We can simplify it by taking the natural logarithm of both the numerator and the denominator separately.

Rewrite the denominator:

x1/x=eln(x)xx^{1/x} = e^{\frac{\ln(x)}{x}}

As xx \to \infty, the exponent ln(x)x\frac{\ln(x)}{x} approaches 0, so:

x1/xe0=1x^{1/x} \to e^0 = 1

Thus, the limit looks like:

limxln(x)x1/x=limxln(x)1=limxln(x)\lim_{x \to \infty} \frac{\ln(x)}{x^{1/x}} = \lim_{x \to \infty} \frac{\ln(x)}{1} = \lim_{x \to \infty} \ln(x)

which diverges to infinity.

Step 2: Apply L'Hopital's Rule

However, we may have made an assumption or overlooked some nuance. Let's go back to applying L'Hopital's Rule, which works in indeterminate forms of type \frac{\infty}{\infty}. The original form is ln(x)x1/x\frac{\ln(x)}{x^{1/x}}, and as we just showed, both the numerator and denominator tend toward infinity, so we can apply L'Hopital’s Rule.

We differentiate the numerator and denominator separately.

  1. Numerator: The derivative of ln(x)\ln(x) is 1x\frac{1}{x}.
  2. Denominator: The derivative of x1/xx^{1/x} requires using the chain rule.

Let’s differentiate x1/x=eln(x)xx^{1/x} = e^{\frac{\ln(x)}{x}}:

ddx(x1/x)=ddx(eln(x)x)=eln(x)xddx(ln(x)x)\frac{d}{dx}(x^{1/x}) = \frac{d}{dx}\left(e^{\frac{\ln(x)}{x}}\right) = e^{\frac{\ln(x)}{x}} \cdot \frac{d}{dx}\left(\frac{\ln(x)}{x}\right)

Now, differentiate ln(x)x\frac{\ln(x)}{x} using the quotient rule:

ddx(ln(x)x)=x1xln(x)x2=1ln(x)x2\frac{d}{dx} \left(\frac{\ln(x)}{x}\right) = \frac{x \cdot \frac{1}{x} - \ln(x)}{x^2} = \frac{1 - \ln(x)}{x^2}

Thus, the derivative of x1/xx^{1/x} is:

ddx(x1/x)=x1/x1ln(x)x2\frac{d}{dx} \left( x^{1/x} \right) = x^{1/x} \cdot \frac{1 - \ln(x)}{x^2}

Now, applying L'Hopital's Rule:

limxln(x)x1/x=limx1xx1/x1ln(x)x2\lim_{x \to \infty} \frac{\ln(x)}{x^{1/x}} = \lim_{x \to \infty} \frac{\frac{1}{x}}{x^{1/x} \cdot \frac{1 - \ln(x)}{x^2}}

As xx \to \infty, this still leads to an indeterminate form, and further simplification or alternative methods might be needed.

Would you like a more in-depth analysis of the behavior of the components, or do you have additional specific parts you'd like me to focus on?

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
L'Hopital's Rule
Exponential Functions
Logarithms

Formulas

lim x->inf ln(x)
L'Hopital's Rule: lim x->inf f(x)/g(x) = lim x->inf f'(x)/g'(x)
x^{1/x} = e^{ln(x)/x}
Derivative of ln(x) = 1/x
Derivative of x^{1/x} = x^{1/x} * (1 - ln(x))/x^2

Theorems

L'Hopital's Rule

Suitable Grade Level

College-level