Problem 19 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0} \frac{\ln \left(x^{2}+1\right)}{x} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Plug in the Limit Value

Start by directly substituting the value of the limit, which is 0, into the expression. This gives us: \[ \lim _{x \to 0} \frac{\ln(x^2 + 1)}{x} = \frac{\ln(0^2 + 1)}{0} = \frac{\ln(1)}{0} = \frac{0}{0}. \]Since this results in the indeterminate form \( \frac{0}{0} \), we cannot determine the limit directly by substitution.

Step 2: Apply L'Hôpital's Rule

Since the expression results in the indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule. This rule states that if the limit of a function \( \frac{f(x)}{g(x)} \) leads to an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit can be computed by differentiating the numerator and the denominator separately.Derivative of the numerator \( \ln(x^2+1) \) is:\[ \frac{d}{dx}[\ln(x^2+1)] = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}. \]Derivative of the denominator \( x \) is:\[ \frac{d}{dx}[x] = 1. \]Thus, applying L'Hôpital's Rule gives us:\[ \lim _{x \rightarrow 0} \frac{\ln(x^2+1)}{x} = \lim_{x \rightarrow 0} \frac{2x}{x^2+1}. \]

Step 3: Compute the New Limit

Now compute the limit after applying L'Hôpital's Rule:\[ \lim _{x \rightarrow 0} \frac{2x}{x^2+1}. \]Substitute \( x = 0 \):\[ \frac{2(0)}{0^2+1} = \frac{0}{1} = 0. \]

Step 4: Conclude the Calculation of the Limit

The limit after applying L'Hôpital's Rule and computing is 0. Since this is a determinate value, it is the calculated limit.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a fantastic tool in calculus for dealing with indeterminate forms, such as when direct substitution in a limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with these forms, the rule tells us we can find the limit of the ratio \( \frac{f(x)}{g(x)} \) by differentiating the numerator and the denominator separately, provided the original limit exists. This means:

• Differentiate the function in the numerator, \( f(x) \).
• Differentiate the function in the denominator, \( g(x) \).
• Take the limit of the resulting quotient of these derivatives.

It's important that the new limit of the derivatives \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) exists, otherwise, further steps are needed. This rule simplifies many complex expressions, turning a seemingly impossible limit into a straightforward problem.

Indeterminate Forms

When calculating limits, there are special scenarios called indeterminate forms that hint a direct approach won't work. Some common forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and more. Essentially, these forms arise from mathematical expressions where the limits of the numerator and denominator or other components are undefined or don't provide clear information.Indeterminate forms don't surrender a limit without more analysis or a different technique. In our example, substituting 0 into the expression \( \frac{\ln(x^2+1)}{x} \) results in \( \frac{0}{0} \), an indeterminate form. That's why a method like L'Hôpital's Rule becomes invaluable to resolve what initially seems unsolvable.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \) is Euler's number. It is widely used in calculus due to its unique properties:

• The natural logarithm's derivative is \( \frac{1}{x} \).
• \( \ln(1) = 0 \) since \( e^0 = 1 \).
• It grows more slowly than linear or exponential functions, making it useful for modeling various natural processes.

In the specific exercise given, \( \ln(x^2 + 1) \) transforms a combination of expressions involving squares and sums into a logarithmic form. This simplification sets the stage for analyzing the function's behavior as \( x \) approaches 0.

Derivatives

Derivatives measure the rate at which a function changes with respect to one of its variables. They are foundational in calculus, allowing for the precise calculation of slopes, rates of change, and limits. To differentiate a complex function, familiar rules like the product rule, quotient rule, and chain rule are used effectively.In the solved limit problem, finding derivatives was key:

• The derivative of \( \ln(x^2 + 1) \) used the chain rule, resulting in \( \frac{2x}{x^2 + 1} \).
• The derivative of \( x \) is straightforwardly \( 1 \).

Applying these derivatives allowed us to transform an indeterminate form into a solvable limit, demonstrating their utility in a dynamic, real-world mathematical setting.

Direct Substitution Method

In calculus, the direct substitution method is often the first step when evaluating limits. Simple and efficient:

• Plug the limit value directly into the expression.
• If the result is a determinate form, the limit is found.
• If it results in an indeterminate form, further methods like factorization or L'Hôpital's Rule are needed.

For the exercise \( \lim _{x \rightarrow 0} \frac{\ln(x^2+1)}{x} \), substituting 0 initially gives \( \frac{0}{0} \), indicating an indeterminate form. Hence, direct substitution shows the need for more powerful techniques to handle the limit properly. It is the straightforward nature of this method that makes it a logical starting point, revealing more complex needed actions.