Problem 33 - University Calculus: Early Transcendentals

Question

Use I'Hópital'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: Evaluate the Limit Form

First, we need to check if the limit is in an indeterminate form. Substitute a small number approaching 0 from the left into both the numerator and the denominator. As \( x \to 0^- \), \( x^2 + 2x \to 0 \) and \( \ln(x^2 + 2x) \to -\infty \), and \( \ln x \to -\infty \). Therefore, the expression is in the indeterminate form \( \frac{-\infty}{-\infty} \).

Step 2: Apply L'Hôpital's Rule

Since the limit is in the form \( \frac{-\infty}{-\infty} \), we can apply L'Hôpital's rule, which states that if the limit of \( \frac{f(x)}{g(x)} \) results in the indeterminate forms \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \), then it equals \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \). Here, take the derivative of the numerator and the denominator:- Derivative of the numerator: \( f(x) = \ln(x^2 + 2x) \). Using the chain rule, \( f'(x) = \frac{1}{x^2 + 2x} \cdot (2x + 2) = \frac{2(x + 1)}{x^2 + 2x} \).- Derivative of the denominator: \( g(x) = \ln x \). The derivative is \( g'(x) = \frac{1}{x} \).

Step 3: Simplify the Expression

Now, using L'Hôpital's rule, the limit becomes:\[\lim_{x \to 0^-} \frac{\ln(x^2 + 2x)}{\ln x} = \lim_{x \to 0^-} \frac{\frac{2(x + 1)}{x^2 + 2x}}{\frac{1}{x}} = \lim_{x \to 0^-} 2(x + 1) \cdot \frac{x}{x^2 + 2x}\]Simplify the expression:\[= \lim_{x \to 0^-} \frac{2x(x + 1)}{x(x + 2)} = \lim_{x \to 0^-} \frac{2(x + 1)}{x + 2}\]

Step 4: Evaluate the Limit

Plug \( x = 0^- \) into the simplified expression:\[= \frac{2(0 + 1)}{0 + 2} = \frac{2}{2} = 1\]Thus, the limit is 1.

Key Concepts

Indeterminate Forms

When evaluating limits, you may encounter expressions that seem unsolvable due to their undefined form. These are known as indeterminate forms. An expression is considered indeterminate when it doesn’t produce a straightforward limit. Examples include forms like \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \). These forms suggest that further work needs to be done to find the limit.

In the given exercise, as we substitute \( x \to 0^- \), the expression \( \frac{\ln(x^2 + 2x)}{\ln x} \) results in \( \frac{-\infty}{-\infty} \), which is a classic indeterminate form. This tells us it's not directly solvable and requires a method like L'Hôpital's Rule to evaluate it. Identifying these forms is crucial because it guides us toward appropriate techniques to solve them.

Derivatives

Derivatives play a central role when dealing with L'Hôpital's Rule as they transform forms like \( \frac{0}{0} \) or \( \frac{\pm\infty}{\pm\infty} \) into ones that are solvable. A derivative measures how a function changes as its input changes, and it’s represented as \( f'(x) \) for the function \( f(x) \).

In this problem, L'Hôpital's Rule is applied by differentiating both the numerator and the denominator. The numerator, \( \ln(x^2 + 2x) \), requires using the chain rule, resulting in \( f'(x) = \frac{2(x + 1)}{x^2 + 2x} \). Meanwhile, the denominator’s derivative, \( \ln x \), simplifies to \( g'(x) = \frac{1}{x} \). By using derivatives, we can simplify the indeterminate form into something manageable that leads us to the final limit.

Limits

The concept of limits addresses how functions behave as they approach a specific value. The limit \( \lim_{x \to c} f(x) \) specifies what happens to \( f(x) \) as \( x \) gets closer and closer to \( c \).

In this exercise, the limit to evaluate is \( \lim_{x \to 0^-} \frac{\ln(x^2 + 2x)}{\ln x} \). Initially, the evaluation leads to an indeterminate form, but after applying L'Hôpital's Rule and substituting the derivatives, we simplify the expression to \( \lim_{x \to 0^-} \frac{2(x+1)}{x+2} \).

Finally, further simplification and evaluation as \( x \to 0^- \) leads to \( \frac{2}{2} = 1 \). This demonstrates how limits help us assess the behavior of expressions near specific points, ultimately providing clear results from initially confusing forms.