Problem 21 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Recognize the Form

As \(x\) approaches 0, both \(\sin(2x)\) and \(\ln(x+1)\) approach 0. This case is an indeterminate form \(\frac{0}{0}\), which can be solved using L'Hôpital's rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's rule states that if \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0}\) or \(\frac{\infty}{\infty}\), the limit can be computed as \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\). Differentiating the numerator and denominator separately gives the derivatives: \(f'(x) = 2\cos(2x)\) and \(g'(x) = \frac{1}{x+1}\).

Step 3: Calculate the Limit After Differentiation

Now, compute the new limit after differentiation: \(\lim_{x \to 0} \frac{2\cos(2x)}{\frac{1}{x+1}} = \lim_{x \to 0} 2(x+1)\cos(2x)\).

Step 4: Evaluate the Limit

Substitute \(x = 0\) in the expression from Step 3: \(2(0+1)\cos(0) = 2 \times 1 \times 1 = 2\).

Key Concepts

Understanding Limits in Calculus

Limits are a fundamental concept in calculus, dealing with the behavior of functions as their inputs approach a certain value. They help us understand what value a function approaches as the input gets arbitrarily close, but not exactly equal, to a specific point. For example, consider the function \(f(x) = \sin(2x)\) as \(x\) approaches \(0\). As \(x\) gets closer and closer to \(0\), \(\sin(2x)\) approaches \(0\) as well. Similarly, \(\ln(x+1)\) approaches \(0\) as \(x\) nears \(0\). However, directly substituting \(0\) into these functions separately would not yield a meaningful answer for a fraction, which is why calculating limits becomes crucial. They help determine the value the expression approaches using the rules and techniques of calculus.
In many cases, especially in the study of calculus, calculating limits is essential for understanding the behavior of functions around points where they are not "well-behaved." This is where advanced techniques, such as L'Hôpital's Rule, come into play to resolve indeterminate forms like \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).

Exploring Indeterminate Forms

Indeterminate forms occur when the limit of a function takes on an unclear or undetermined shape, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms make it challenging to ascertain the limit directly. In our exercise, as \(x\rightarrow 0\), we noticed the expression \(\frac{\sin(2x)}{\ln(x+1)}\) resulted in \(\frac{0}{0}\), which is an indeterminate form.
To resolve such indeterminate forms, we can't merely substitute values directly, so we use methods like L'Hôpital's Rule. This allows us to differentiate the numerator and the denominator separately to find a new limit that might be determinate. By handling the indeterminate form correctly, we can find out what value is approached, ultimately offering a deeper understanding of the function's behavior near the point of interest.

Focusing on Trigonometric Limits

Trigonometric limits often appear in calculus problems, especially when dealing with functions that oscillate, like sine and cosine. These functions have well-known limit properties that are extensively utilized. For instance, the standard trigonometric limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) is fundamental for solving many calculus problems. It tells us that when \(x\) approaches zero, the quotient of \(\sin(x)\) and \(x\) approaches 1.
In the context of our exercise, \(\sin(2x)\) as \(x\rightarrow 0\) turns into a classic trigonometric problem. Using L'Hôpital's Rule, we differentiate \(\sin(2x)\) to simplify and solve the limit. Trigonometric limits such as these are widespread, and understanding their derivations and applications is a crucial skill for anyone studying calculus. They provide insights not only in exercises but also in understanding waves and periodic functions that underpin much of physics and engineering.

Derivatives for Limit Calculation

Derivatives play a pivotal role in calculating limits, particularly when dealing with indeterminate forms. As seen in our solution, L'Hôpital's Rule involves using the derivatives of the functions in the numerator and the denominator to resolve an indeterminate form. By differentiating \(\sin(2x)\) to get \(2\cos(2x)\) and \(\ln(x+1)\) to get \(\frac{1}{x+1}\), the original limit transforms into something more manageable.
Derivatives essentially give us the rate of change or the slope of a curve at any point, which is paramount for finding limits in complex cases. With L'Hôpital's Rule, after taking the derivative of the numerator and the denominator, you re-evaluate the limit, often simplifying the problem and making it easier to solve. Understanding how and when to apply derivatives in the context of calculating limits is a crucial skill in calculus, aiding in the solving of complex limit problems skillfully and accurately.