Problem 10 - Calculus Single Variable

Question

Use l'Hôpital's Rule to find the limit, if it exists. \(\lim _{x \rightarrow-1} \frac{x+x^{2}}{\ln (2+x)}\)

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Check if l'Hôpital's Rule is Applicable

To use l'Hôpital's Rule, the limit must initially give an indeterminate form, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Substitute \(x = -1\) into the expression: \(\frac{-1 + (-1)^2}{\ln(2-1)} = \frac{0}{\ln(1)} = \frac{0}{0}\). Since it results in \(\frac{0}{0}\), l'Hôpital's Rule can be applied.

Step 2: Differentiate the Numerator and Denominator

Find the derivative of the numerator, \(x + x^2\). The derivative is \(1 + 2x\). Find the derivative of the denominator, \(\ln(2+x)\). The derivative is \(\frac{1}{2+x}\).

Step 3: Apply l'Hôpital's Rule

According to l'Hôpital's Rule, take the limit of the derivatives: \(\lim_{x \to -1} \frac{1 + 2x}{\frac{1}{2+x}}\). Simplify this to \(\lim_{x \to -1} (1 + 2x)(2 + x)\).

Step 4: Simplify and Calculate the New Limit

Expand \((1 + 2x)(2 + x)\) to obtain \(2 + x + 4x + 2x^2 = 2 + 5x + 2x^2\). Substitute \(x = -1\) into the expression: \(2 + 5(-1) + 2(-1)^2 = 2 - 5 + 2 = -1\).

Step 5: Conclude the Calculation

The simplified limit evaluation gives \(-1\). This is the value of the limit as \(x\) approaches \(-1\).

Key Concepts

Limit Calculation

Limits define the approaching behavior of a function as the input approaches a certain value. In this problem, we consider the limit as \( x \) approaches \(-1\). The task is to determine the behavior of the function \( \frac{x + x^2}{\ln(2 + x)} \). For a limit to be applicable for l'Hôpital's Rule, the direct substitution of \( x = -1 \) must result in an indeterminate form. Here, substituting \(-1\) into the function gives \( \frac{0}{0} \), which is an indeterminate form—a requirement for employing l'Hôpital's Rule. Thus, the initial step in calculating this limit confirms the applicability of the rule. Through each step in this limit calculation, understanding both the need for and application of l'Hôpital's Rule is essential for finding the answer.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. Essential in applying l'Hôpital's Rule is understanding how to calculate derivatives, as the rule requires differentiating both the numerator and denominator of the given function.In our exercise, the numerator \( x + x^2 \) differentiates to \( 1 + 2x \). This process involves applying simple polynomial differentiation rules:

• The derivative of \( x \) is \( 1 \).
• The derivative of \( x^2 \) is \( 2x \).

For the denominator, \( \ln(2+x) \), the derivative calculation utilizes the chain rule, resulting in \( \frac{1}{2+x} \). This involves acknowledging the derivative of a natural logarithm and employing the chain rule since the expression inside the logarithm requires differentiation.These derivatives are crucial as they simplify the original limit, allowing further steps using the l'Hôpital's Rule.

Indeterminate Forms

Indeterminate forms arise when limits don't immediately resolve to a clear value, often appearing as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the context of limits, an indeterminate form means simply substituting a value won't directly lead to the answer because the expression's behavior isn't easily resolved until further evaluation.In this exercise, plugging \( x = -1 \) into \( \frac{x + x^2}{\ln(2 + x)} \) results in \( \frac{0}{0} \), a typical indeterminate form. Identifying an indeterminate form is crucial for deciding that l'Hôpital's Rule should be applied.l'Hôpital's Rule then provides the technique to reevaluate these forms by differentiating the numerator and denominator separately, simplifying the expression to a point where the limit can be calculated directly.

Numerator and Denominator Differentiation

Differentiating both the numerator and the denominator forms the core operation in l'Hôpital's application. Here is where you translate the indeterminate form into a more manageable setup for limit calculation.In our example, the function given creates the indeterminate form \( \frac{0}{0} \) after substituting \( x = -1 \). By differentiating the numerator \( x + x^2 \) to obtain \( 1 + 2x \) and the denominator \( \ln(2 + x) \) to derive \( \frac{1}{2+x} \), the expression is transformed.The resulting limit \[\lim_{x \to -1} \frac{1 + 2x}{\frac{1}{2+x}}\]can then be further simplified into \[\lim_{x \to -1} (1 + 2x)(2 + x)\]This shows how differentiation turns complex problems into manageable steps that facilitate precise limit evaluation. The careful handling of differentiation will solidify understanding and make applying l'Hôpital's Rule more intuitive.