Problem 36 - Calculus: Early Transcendentals

Question

Determine the infinite limit. \( \displaystyle \lim_{x \to 0^+}\ln (\sin x) \)

Step-by-Step Solution

Answer

The infinite limit is \(-\infty\).

Step 1: Substitute and Analyze

To find the limit as \( x \to 0^+ \), we need to substitute values approaching 0 from the positive side into \( \sin x \). Recall that the sine function approaches 0 as \( x \to 0^+ \), specifically: \( \sin x \to 0^+ \).

Step 2: Evaluate the Logarithm

The natural logarithm function, \( \ln(x) \), approaches negative infinity as \( x \) approaches 0 from the positive side. Therefore, since \( \sin x \to 0^+ \), \( \ln(\sin x) \to -\infty \).

Step 3: Conclusion

Putting the previous analysis together, we find that as \( x \to 0^+ \), \( \ln(\sin x) \to -\infty \). Thus, the infinite limit is \( -\infty \).

Key Concepts

Natural Logarithm

The natural logarithm, noted as \( \ln(x) \), is a fundamental function in mathematics, known for its unique properties. Essentially, it's the inverse operation of the exponential function \( e^x \).
When you have \( \ln(e^x) \), you get \( x \) back. This relationship is what makes the natural logarithm so useful in various applications, including calculus, algebra, and beyond.One essential property of \( \ln(x) \) is its behavior as \( x \) approaches 0 from the positive side.
In such cases, \( \ln(x) \) approaches negative infinity. This is because you're essentially asking "to what power must \( e \) be raised to produce a tiny number close to zero?"
The answer is a very large negative power, reflecting that \( \ln(x) \to -\infty \) as \( x \to 0^+ \).
This property is critical when dealing with limits in calculus, especially when you explore how functions behave near points where they might not be defined.When taking the limit of \( \ln(\sin x) \) as \( x \to 0^+ \), it utilizes this logarithmic property since \( \sin x \to 0^+ \).
Hence, the outcome of \( \ln(\sin x) \to -\infty \) as \( x \to 0^+ \) makes sense due to the nature of the logarithm itself.

Sine Function

The sine function is one of the fundamental trigonometric functions, typically represented as \( \sin(x) \).
It describes a wave-like pattern that oscillates between -1 and 1, with a period of \( 2\pi \). This functionality makes it important in various fields such as physics, engineering, and signal processing.As \( x \) approaches 0, \( \sin(x) \) behaves in a particular way that influences many problems in calculus and beyond. Near zero, \( \sin(x) \) is approximately equal to \( x \) itself when measured in radians, i.e., \( \sin(x) \approx x \).
Therefore, when \( x \to 0^+ \), \( \sin(x) \to 0^+ \). This characteristic is crucial when evaluating limits, as shown in the exercise where the limit of \( \ln(\sin x) \) needed to consider \( \sin x \to 0^+ \).Understanding how sine behaves as inputs get small is powerful for solving many calculus problems. By approximating \( \sin(x) \) with \( x \), you have a simple yet effective method for evaluating limits and derivates involving trigonometric functions.

Limits in Calculus

Limits are at the core of calculus; they describe the behavior of functions as inputs get extremely close to a particular point. A foundational concept in calculus, limits help understand how functions behave at boundaries or points of discontinuity.Take the example of \( \lim_{x \to 0^+}\ln(\sin(x)) \).
Here, establishing that \( \sin(x) \to 0^+ \) simplifies the process, as the natural logarithm's tendency to \(-\infty\) when approaching zero can be applied.
Hence, the limit becomes \(-\infty\), reflecting the behavior of the composite function as \( x \to 0^+ \).Working with limits requires careful consideration of the direction from which you approach a point, noted as \( x \to a^+ \) for approaching from the right, or \( x \to a^- \) from the left.
These distinctions guide us in understanding different behaviors that might not be evident otherwise. The step-by-step analysis of limits also often involves substitution and approximation, essential techniques in solving complex calculus problems.