Question

as $x\to\infty$. explain.

1. comparing logarithm and power functions

(a) writing to learn show that $\ln x$ grows slower than $x^{1/n}$ as $x\to\infty$ for any positive integer $n$, even $n = 1,000,000$. explain.
(b) writing to learn show that for any number $a > 0$, $\ln x$ grows slower than $x^a$ as $x\to\infty$. explain.

Explanation:

Step1: Recall the limit definition for growth rates

To determine if \( \ln x \) grows slower than \( x^{1/n} \) (or \( x^a \)) as \( x \to \infty \), we use the limit \( \lim_{x \to \infty} \frac{\ln x}{x^k} \) where \( k=\frac{1}{n} \) (or \( k = a>0 \)). If this limit is \( 0 \), then \( \ln x \) grows slower than \( x^k \).

Step2: Apply L'Hospital's Rule

For \( \lim_{x \to \infty} \frac{\ln x}{x^k} \) (where \( k>0 \)), since as \( x \to \infty \), \( \ln x \to \infty \) and \( x^k \to \infty \), we can apply L'Hospital's Rule. The derivative of the numerator \( \ln x \) is \( \frac{1}{x} \), and the derivative of the denominator \( x^k \) is \( kx^{k - 1} \). So the limit becomes:

$$ \lim_{x \to \infty} \frac{\frac{1}{x}}{kx^{k - 1}}=\lim_{x \to \infty} \frac{1}{kx^{k}} $$

Step3: Evaluate the new limit

As \( x \to \infty \), \( x^k \to \infty \) (because \( k>0 \)). So \( \frac{1}{kx^{k}} \to 0 \) as \( x \to \infty \).

Part (a) Specifics:

For \( k=\frac{1}{n} \) where \( n \) is a positive integer (even \( n = 1000000 \), so \( k=\frac{1}{1000000}>0 \)), the limit \( \lim_{x \to \infty} \frac{\ln x}{x^{1/n}}=\lim_{x \to \infty} \frac{1}{\frac{1}{n}x^{1/n}} = 0 \) (since \( x^{1/n} \to \infty \) as \( x \to \infty \)). Thus, \( \ln x \) grows slower than \( x^{1/n} \).

Part (b) Specifics:

For any \( a>0 \), let \( k = a \). Then \( \lim_{x \to \infty} \frac{\ln x}{x^a}=\lim_{x \to \infty} \frac{1}{ax^{a}} = 0 \) (since \( x^a \to \infty \) as \( x \to \infty \) for \( a>0 \)). So \( \ln x \) grows slower than \( x^a \).

Answer:

To show \( \ln x \) grows slower than \( x^{1/n} \) (or \( x^a \)) as \( x \to \infty \):

1. Use the limit \( \lim_{x \to \infty} \frac{\ln x}{x^k} \) ( \( k=\frac{1}{n}>0 \) or \( k = a>0 \) ).
2. Apply L'Hospital's Rule (since \( \ln x \to \infty \) and \( x^k \to \infty \) as \( x \to \infty \)):

• Derivative of numerator: \( \frac{d}{dx}(\ln x)=\frac{1}{x} \)
• Derivative of denominator: \( \frac{d}{dx}(x^k)=kx^{k - 1} \)
• New limit: \( \lim_{x \to \infty} \frac{\frac{1}{x}}{kx^{k - 1}}=\lim_{x \to \infty} \frac{1}{kx^{k}} \)

1. Evaluate: As \( x \to \infty \), \( x^k \to \infty \) ( \( k>0 \) ), so \( \frac{1}{kx^{k}} \to 0 \).

Thus, \( \lim_{x \to \infty} \frac{\ln x}{x^k}=0 \), meaning \( \ln x \) grows slower than \( x^k \) ( \( x^{1/n} \) or \( x^a \) ) as \( x \to \infty \).