Understanding Continuity of a Function

For a function \( f(x) \) to be continuous at a specific point, say \( x = a \), three conditions must be met:

1. The function must be defined at \( x = a \), meaning \( f(a) \) exists.
2. The limit of the function as \( x \) approaches \( a \) must exist, meaning \( \lim_{x \to a} f(x) \) exists.
3. The limit of the function as \( x \) approaches \( a \) must be equal to the function's value at \( a \), i.e., \( \lim_{x \to a} f(x) = f(a) \).

In this question, we are given a piecewise function \( f(x) \) that is said to be continuous at \( x = \frac{\pi}{4} \). The function is defined as:

\[ f(x) = \begin{cases} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases} \]

The point of interest is \( a = \frac{\pi}{4} \). The value of the function at this point is given as \( f(\frac{\pi}{4}) = k \).

For continuity at \( x = \frac{\pi}{4} \), the third condition must hold: \( \lim_{x \to \frac{\pi}{4}} f(x) = f(\frac{\pi}{4}) \).

So, we need to find the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{4} \) and set it equal to \( k \).

\[ k = \lim_{x \to \frac{\pi}{4}} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x} \]

Evaluating the Limit for Continuity

Let's evaluate the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x} \).

If we directly substitute \( x = \frac{\pi}{4} \) into the expression, we get:

Numerator: \( \tan(\frac{\pi}{4} - \frac{\pi}{4}) = \tan(0) = 0 \)

Denominator: \( \cot(2 \times \frac{\pi}{4}) = \cot(\frac{\pi}{2}) = 0 \)

This gives us the indeterminate form \( \frac{0}{0} \). We can use either trigonometric identities or L'Hopital's rule to evaluate this limit.

Method 1: Using Trigonometric Identities

Recall the identities: \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \) and \( \cot(2A) = \frac{1 - \tan^2 A}{2 \tan A} \).

Using \( A = \frac{\pi}{4} \) and \( B = x \), we have \( \tan(\frac{\pi}{4} - x) = \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x} \).

Using \( A = x \), we have \( \cot(2x) = \frac{1}{\tan(2x)} = \frac{1}{\frac{2 \tan x}{1 - \tan^2 x}} = \frac{1 - \tan^2 x}{2 \tan x} \).

So, the expression becomes:

\[ \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x} = \frac{\frac{1 - \tan x}{1 + \tan x}}{\frac{1 - \tan^2 x}{2 \tan x}} \]

\[ = \frac{1 - \tan x}{1 + \tan x} \times \frac{2 \tan x}{1 - \tan^2 x} \]

Since \( 1 - \tan^2 x = (1 - \tan x)(1 + \tan x) \), we get:

\[ = \frac{1 - \tan x}{1 + \tan x} \times \frac{2 \tan x}{(1 - \tan x)(1 + \tan x)} \]

For \( x \neq \frac{\pi}{4} \), \( \tan x \neq 1 \), so we can cancel the \( (1 - \tan x) \) term:

\[ = \frac{2 \tan x}{(1 + \tan x)(1 + \tan x)} = \frac{2 \tan x}{(1 + \tan x)^2} \]

Now, evaluate the limit as \( x \to \frac{\pi}{4} \). As \( x \to \frac{\pi}{4} \), \( \tan x \to \tan \frac{\pi}{4} = 1 \).

\[ \lim_{x \to \frac{\pi}{4}} \frac{2 \tan x}{(1 + \tan x)^2} = \frac{2 \times 1}{(1 + 1)^2} = \frac{2}{2^2} = \frac{2}{4} = \frac{1}{2} \]

Method 2: Using L'Hopital's Rule

Since we have the \( \frac{0}{0} \) indeterminate form, we can apply L'Hopital's rule. We differentiate the numerator and the denominator with respect to \( x \).

Derivative of the numerator \( \tan(\frac{\pi}{4} - x) \):

Using the chain rule, \( \frac{d}{du} (\tan u) = \sec^2 u \) and \( \frac{d}{dx} (\frac{\pi}{4} - x) = -1 \). So, \( \frac{d}{dx} \tan(\frac{\pi}{4} - x) = \sec^2(\frac{\pi}{4} - x) \times (-1) = -\sec^2(\frac{\pi}{4} - x) \).

Derivative of the denominator \( \cot(2x) \):

Using the chain rule, \( \frac{d}{du} (\cot u) = -\csc^2 u \) and \( \frac{d}{dx} (2x) = 2 \). So, \( \frac{d}{dx} \cot(2x) = -\csc^2(2x) \times 2 = -2\csc^2(2x) \).

Now, apply L'Hopital's rule:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\tan (\frac{\pi}{4} - x)}{\cot 2x} = \lim_{x \to \frac{\pi}{4}} \frac{-\sec^2(\frac{\pi}{4} - x)}{-2\csc^2(2x)} \]

\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sec^2(\frac{\pi}{4} - x)}{2\csc^2(2x)} \]

Now, substitute \( x = \frac{\pi}{4} \) into the derivatives:

\( \sec^2(\frac{\pi}{4} - \frac{\pi}{4}) = \sec^2(0) = (\frac{1}{\cos 0})^2 = (\frac{1}{1})^2 = 1 \).

\( \csc^2(2 \times \frac{\pi}{4}) = \csc^2(\frac{\pi}{2}) = (\frac{1}{\sin \frac{\pi}{2}})^2 = (\frac{1}{1})^2 = 1 \).

So, the limit is:

\[ \frac{1}{2 \times 1} = \frac{1}{2} \]

Both methods yield the same limit value, which is \( \frac{1}{2} \).

Finding the Value of k for Continuity

For the function \( f(x) \) to be continuous at \( x = \frac{\pi}{4} \), the limit as \( x \) approaches \( \frac{\pi}{4} \) must equal the function value at \( x = \frac{\pi}{4} \).

So, \( \lim_{x \to \frac{\pi}{4}} f(x) = f(\frac{\pi}{4}) \).

We found that \( \lim_{x \to \frac{\pi}{4}} f(x) = \frac{1}{2} \), and we are given that \( f(\frac{\pi}{4}) = k \).

Therefore, we must have \( k = \frac{1}{2} \).

The value of \( k \) that makes the function continuous at \( x = \frac{\pi}{4} \) is \( \frac{1}{2} \).

Revision Table: Key Continuity and Limit Concepts

Additional Information: Related Concepts in Calculus

Understanding continuity and limits is fundamental in calculus. Here are a few more related concepts:

Connecting Limits and Continuity

The definition of continuity \( \lim_{x \to a} f(x) = f(a) \) directly links the concept of a limit to the concept of continuity. If the limit exists but doesn't equal the function value (or the function value isn't defined), the function is discontinuous at that point. The limit describes the behavior of the function *near* the point, while \( f(a) \) describes the behavior *at* the point.

Derivability and Continuity

An important theorem states that if a function is differentiable at a point, then it must be continuous at that point. However, the converse is not true; a function can be continuous at a point but not differentiable (a classic example is \( f(x) = |x| \) at \( x=0 \)). This means continuity is a necessary but not sufficient condition for differentiability.

One-Sided Limits

For the limit \( \lim_{x \to a} f(x) \) to exist, the left-hand limit \( \lim_{x \to a^-} f(x) \) and the right-hand limit \( \lim_{x \to a^+} f(x) \) must both exist and be equal. For a function to be continuous at \( a \), this common limit value must also equal \( f(a) \).

For piecewise functions like the one in the question, checking continuity at the points where the definition changes often involves evaluating one-sided limits or, as done here, the overall limit using the definition for \( x \neq a \).