Problem 17 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow \infty} \frac{(\ln x)^{3}}{x^{2}} $$

Step-by-Step Solution

Answer

The given limit converges to 0 after applying l'Hôpital's Rule twice and the final expression is, $$ \lim _{x \rightarrow \infty} \frac{6(\ln x) - 3(\ln x)^2}{4x^3} = 0 $$

Step 1: 2. Differentiating the numerator and the denominator

Now let's find the derivatives of both the numerator and the denominator: Numerator derivative \((\ln x)^3\): Using the chain rule, we have: \( \frac{d}{dx} (\ln x)^3 = 3(\ln x)^2 \frac{d}{dx} (\ln x) \) Since the derivative of \( \ln x \) is \( \frac{1}{x} \), the derivative of the numerator is: \( 3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x} \) Denominator derivative \(x^2\): \( \frac{d}{dx} x^2 = 2x \)

Step 2: 3. Apply l'Hôpital's Rule

Since we have an indeterminate form of the type ∞/∞, let's apply l'Hôpital's Rule. To do this, we need to compute the limit of the ratio of the derivatives obtained above: $$ \lim _{x \rightarrow \infty} \frac{\frac{3(\ln x)^2}{x}}{2x} $$

Step 3: 4. Simplifying the expression

Now let's simplify the expression by dividing the numerator by the denominator: $$ \lim _{x \rightarrow \infty} \frac{3(\ln x)^2}{x} \cdot \frac{1}{2x} = \lim _{x \rightarrow \infty} \frac{3(\ln x)^2}{2x^2} $$

Step 4: 5. Apply l'Hôpital's Rule again

Again, the expression has an indeterminate form of the type ∞/∞. So, we can apply l'Hôpital's Rule a second time. First, let us find the derivatives of the new numerator and denominator: New numerator derivative \(\frac{3(\ln x)^2}{x}\): Using the chain rule and the quotient rule, we get: \( \frac{d}{dx} \frac{3(\ln x)^2}{x} = \frac{6(\ln x)(\frac{1}{x})\cdot x - 3(\ln x)^2 (\frac{1}{x})}{x^2} = \frac{6(\ln x) - 3(\ln x)^2}{x^2} \) New denominator derivative \(2x^2\): \( \frac{d}{dx} 2x^2 = 4x \) Now, let's apply l'Hôpital's Rule: $$ \lim _{x \rightarrow \infty} \frac{\frac{6(\ln x) - 3(\ln x)^2}{x^2}}{4x} $$

Step 5: 6. Simplify the expression again

Simplify the expression by dividing the numerator by the denominator: $$ \lim _{x \rightarrow \infty} \frac{6(\ln x) - 3(\ln x)^2}{x^2} \cdot \frac{1}{4x} = \lim _{x \rightarrow \infty} \frac{6(\ln x) - 3(\ln x)^2}{4x^3} $$ In this case, as x approaches infinity, the terms \(\frac{6(\ln x)}{4x^3}\) and \(\frac{3(\ln x)^2}{4x^3}\) both approach 0. Hence, the limit is: $$ \lim _{x \rightarrow \infty} \frac{6(\ln x) - 3(\ln x)^2}{4x^3} = 0 $$ So, the given limit converges to 0.

Key Concepts

Limit Evaluation

Evaluating limits is a fundamental concept in calculus, vital for understanding the behavior of functions as they approach certain points. Limits help in predicting the value that a function approaches as the input grows infinitely large or small. In particular, the limit \( \lim_{x \to \infty} \frac{(\ln x)^3}{x^2} \) describes how the function behaves as \( x \) becomes very large.

To evaluate this limit, we first check if the function forms an indeterminate expression, such as \( \frac{\infty}{\infty} \), which it does. This opens the door to using powerful tools like l'Hôpital's Rule to find the result.

Remember, limits allow us to make sense of functions at points where they might not even be defined explicitly.

Indeterminate Forms

Indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), arise in calculus when directly substituting values into functions gives an undefined outcome. These require us to do more than simply plug in numbers. We need to perform additional steps to resolve these limits.

In the scenario of \( \lim_{x \to \infty} \frac{(\ln x)^3}{x^2} \), the expression takes the form \( \frac{\infty}{\infty} \), indicating a need for l'Hôpital's Rule. This rule guides us towards differentiating the numerator and denominator separately until a determined form is achieved.

Indeterminate forms require special techniques because they imply regular substitution isn't enough, urging for analytical manipulation instead.

Chain Rule

When differentiating composite functions, the chain rule is crucial. It allows us to find the derivative of functions composed of other functions, such as \( (\ln x)^3 \).

The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). Applying it to \( (\ln x)^3 \), we differentiate by treating \( \ln x \) as the inner function and 3 as the exponent. The derivative becomes \( 3(\ln x)^2 \cdot \frac{1}{x} \), where \( \frac{1}{x} \) is the derivative of \( \ln x \).

This tool is indispensable for decomposition in differentiation, especially when handling functions nested within each other.

Quotient Rule

The quotient rule is used to differentiate functions that are ratios of two other functions, essential in solving limits involving fractions.

In the expression \( \frac{3(\ln x)^2}{x} \), the quotient rule requires differentiating the numerator and denominator separately. The formula is \( \frac{g'(x)f(x) - f'(x)g(x)}{(g(x))^2} \), where \( f(x) \) is the numerator and \( g(x) \) is the denominator.

After differentiating both parts, simplify the derivatives to continue evaluating using l'Hôpital's Rule repeatedly, as needed. Understanding and applying the quotient rule help in breaking down complex fraction derivatives, making progression through problems smoother.