Problem 19 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x) \quad\\{\text { Ans. } 0\\} $$

Step-by-Step Solution

Answer

The limit of the given function \(\lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x)\) can be found by first evaluating the limit of the function inside the parenthesis (\(\sin \pi x - \cos \pi x\)), which results in 1. Next, we evaluate the limit of the natural logarithm function with this value, which gives us \(\ln(1) = 0\). Thus, the final answer is \(0\).

Step 1: Evaluate the limit of sine and cosine functions separately

Since our main function is the natural logarithm of the function inside the parenthesis, we can start by evaluating the limit of the function inside the parenthesis, which involves sine and cosine functions: \[ \lim _{x \rightarrow 1}(\sin \pi x - \cos \pi x) \] Now, as x approaches 1, we can replace x with 1 in the sine and cosine parts: \[ \sin \pi (1) - \cos \pi(1) = \sin \pi - \cos \pi \] We know that \(\sin \pi = 0\) and \(\cos \pi = -1\). Therefore, after replacing with these values: \[ 0 - (-1) = 1 \]

Step 2: Evaluate the limit of the natural logarithm function

Now that we have found the limit of the function inside the parenthesis, we can evaluate the limit of the natural logarithm function with this value: \[ \lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x) = \ln (1) \] We know that the natural logarithm of 1 is 0: \[ \ln (1) = 0 \]

Step 3: Final answer

Combining the results of Steps 1 and 2, we have found the limit of the given function: \[ \lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x) = 0 \] Thus, the final answer is 0.

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm whose base is the irrational number \( e \), approximately equal to 2.71828. It is used frequently in calculus, especially when working with exponential growth and decay or evaluating limits that involve exponential functions.
The natural logarithm of 1 is always 0. This is because \( e^0 = 1 \). Understanding the properties of the natural logarithm can simplify complex expressions and help solve problems involving exponential and logarithmic functions.
In limits, the natural logarithm can transform complex limit expressions into easier forms, allowing for simpler evaluations. Knowing that \( \ln(1) = 0 \) is a fundamental principle, crucial for solving many calculus problems efficiently.

Trigonometric Limits

Trigonometric limits are important for solving calculus problems involving trigonometric functions like sine, cosine, and tangent. They often involve evaluating the behavior of these functions as the variable approaches a certain point.
In the context of this exercise, we deal with the functions \( \sin(\pi x) \) and \( \cos(\pi x) \) as \( x \) approaches 1. Recognizing how these functions behave at key angles, like \( \pi \), helps evaluate limits more effectively. For example:

• \( \sin(\pi) = 0 \)
• \( \cos(\pi) = -1 \)

By substituting these values, we reduce the problem to simpler arithmetic, allowing us to find the limit quickly.
Mastery of trigonometric limits aids in solving complex limit problems, as these functions frequently appear in various forms in calculus.

Evaluation of Limits

Evaluating limits involves determining the value that a function approaches as the input approaches a certain specific value. It is a core concept in calculus, providing groundwork for derivative and integral concepts.
The key steps include:

• Substitute the value directly into the function, if possible.
• Simplify the function to identify indeterminate forms, like \( \frac{0}{0} \), and apply algebraic techniques to resolve these forms.
• Use trigonometric identities or logarithmic properties to aid in simplification.

In our example, evaluating \( \lim _{x \rightarrow 1} (\sin \pi x - \cos \pi x) \) simplifies by using known sine and cosine values to yield 1. Further, this helps in evaluating \( \ln(1) = 0 \).
Evaluating limits is crucial in approximating functions and forms the basis for more complex calculus concepts.

Sine and Cosine Functions

Sine and cosine are fundamental trigonometric functions representing the coordinates on the unit circle. These functions repeat values cyclically, known as periodicity, with periods of \( 2\pi \).
These functions are defined as:

• \( \sin(\theta) = \frac{opposite}{hypotenuse} \)
• \( \cos(\theta) = \frac{adjacent}{hypotenuse} \)

In the unit circle, \( \sin(\theta) \) represents the y-coordinate, and \( \cos(\theta) \) represents the x-coordinate.
When dealing with limits, recognizing values of sine and cosine at key angles (such as \( 0, \pi/2, \pi, 3\pi/2, 2\pi \)) helps in quickly evaluating and simplifying expressions. In our limit evaluation \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \), substituting these simplifies the problem to easier arithmetic.
Mastering these functions is vital for tackling various problems in calculus, physics, and engineering.