Problem 33 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{x \rightarrow 0^{+}} \frac{\ln \left(x^{2}+2 x\right)}{\ln x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Verify Indeterminate Form

We first check whether the given limit \( \lim _{x \rightarrow 0^{+}} \frac{\ln \left(x^{2}+2x\right)}{\ln x} \) is in an indeterminate form. - As \( x \to 0^+ \), \( \ln(x) \to -\infty \).- For \( \ln(x^2 + 2x) \) as \( x \rightarrow 0^+ \), \( x^{2} + 2x \to 0 \) makes \( \ln(x^2 + 2x) \to -\infty \).- Therefore, the limit \( \frac{\ln \left(x^{2}+2x\right)}{\ln x} \) is indeed in the indeterminate form \( \frac{-\infty}{-\infty} \), and l'Hôpital's Rule can be applied.

Step 2: Differentiate Numerator and Denominator

To apply l'Hôpital's Rule, we differentiate the numerator and the denominator separately:- The derivative of the numerator, \( \ln(x^2 + 2x) \), is \( \frac{1}{x^2 + 2x} \cdot (2x + 2) = \frac{2(x + 1)}{x^2 + 2x} \).- The derivative of the denominator, \( \ln(x) \), is \( \frac{1}{x} \).

Step 3: Apply l’Hôpital’s Rule

Apply l'Hôpital's Rule using the derivatives you just found:\[\lim _{x \rightarrow 0^{+}} \frac{\ln \, (x^{2}+2x)}{\ln \, x} = \lim _{x \rightarrow 0^{+}} \frac{\frac{2(x+1)}{x^2 + 2x}}{\frac{1}{x}} = \lim _{x \rightarrow 0^{+}} \frac{2(x+1) \cdot x}{x^2 + 2x} = \lim _{x \rightarrow 0^{+}} \frac{2x(x+1)}{x(x+2)}\]

Step 4: Simplify the Expression

Simplify the expression by canceling common terms:\[\lim _{x \rightarrow 0^{+}} \frac{2x(x+1)}{x(x+2)} = \lim _{x \rightarrow 0^{+}} \frac{2(x+1)}{x+2}\] - As \( x \to 0^+ \), this simplifies to \( \frac{2(1)}{2} = 1 \).

Step 5: Conclude the Limit

After simplifying, the limit results in a determinate value:- \( \lim _{x \rightarrow 0^{+}} \frac{\ln \left(x^{2}+2x\right)}{\ln x} = 1 \).

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms are expressions that arise when evaluating limits and do not yield a clear answer. Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 0^0 \), \( 1^{\infty} \), and \( \infty^0 \)

These forms require additional analysis to determine the actual limit value, if it exists.
In the original exercise, we evaluated the limit \( \lim _{x \rightarrow 0^{+}} \frac{\ln (x^{2}+2x)}{\ln x} \), which initially results in \( \frac{-\infty}{-\infty} \). This is a classic indeterminate form that can be managed using l'Hôpital's Rule.
Understanding indeterminate forms is crucial because they appear frequently in mathematics, especially when dealing with complex functions in calculus.

Limits in Calculus

In calculus, limits help us understand the behavior of functions as they approach certain points. Limits can describe what a function looks like near a point, even if the function doesn't exist at that point.

There are some key points about limits:

• A limit approaching from the right is denoted as \( x \rightarrow 0^{+} \), and from the left as \( x \rightarrow 0^{-} \).
• If a function approaches a specific value, we can find the limit at that point.
• If a function's value doesn't stabilize to a number, the limit may not exist.

In this exercise, we looked at how to handle limits when dealing with complex functions like logarithmic expressions. Limits can be evaluated directly, but for indeterminate forms, we often require techniques like l'Hôpital’s Rule. Recognizing when and how to apply these techniques is a vital skill in calculus.

Logarithmic Differentiation

Logarithmic differentiation is a method used to differentiate functions that are challenging to handle with other techniques. It's particularly useful when dealing with products, quotients, or powers of functions.

Here's how logarithmic differentiation is useful:

• It simplifies complex functions where direct differentiation is cumbersome.
• Use it when the base of exponential functions variable.
• It combines the power rule, product rule, and chain rule.

In the provided exercise, we used logarithmic differentiation to find the derivative of \( \ln(x^2 + 2x) \). By applying this technique, we successfully broke down the differentiation into manageable parts.
This simplifies the process of applying l'Hôpital’s Rule afterward. The technique enhances derivative calculations, streamlining approaches for limits involving logarithmic functions.