Question

Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

• A. \(0\)
• B. \(-2\)
• C. \(-1\)
• D. \(2\)

Hint:

When computing limits containing basic trigonometric expressions at $0$, look for standard identities like \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) or \(\lim_{x \to 0} \frac{\tan x}{x} = 1\) to simplify the expression immediately.

Answer 1

The Correct Option is D

Concept: For a function \(f(x)\) to be mathematically continuous at a specific point \(x = a\), it must fulfill three strict foundational conditions simultaneously:
• The functional value \(f(a)\) must be well-defined at that coordinate location.
• The limit of the function as \(x\) approaches \(a\) must exist: \(\lim_{x \to a} f(x) = L\).
• The computed limit value must be exactly equal to the defined functional value: \[ \lim_{x \to a} f(x) = f(a) \] We will use the standard fundamental limit theorem for trigonometric functions: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Step 1: Identify the functional value at the origin
From the given piecewise function definition, when \(x = 0\), the output value is explicitly defined as: \[ f(0) = k \quad \cdots (1) \]

Step 2: Evaluate the limit value as \(x \to 0\)
We evaluate the limit of \(f(x)\) as \(x\) approaches $0$ using the expression for \(x \neq 0\): \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right) \] Using sum properties of limit arithmetic, we can split this into two separate simple limit expressions: \[ \lim_{x \to 0} f(x) = \left( \lim_{x \to 0} \frac{\sin x}{x} \right) + \left( \lim_{x \to 0} \cos x \right) \] We evaluate each component based on standard limits and direct evaluation:
• \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)
• \(\lim_{x \to 0} \cos x = \cos(0) = 1\) Adding these two limits together gives: \[ \lim_{x \to 0} f(x) = 1 + 1 = 2 \quad \cdots (2) \]

Step 3: Equate the limit to the functional value for continuity
Since the problem states that \(f(x)\) is continuous at \(x = 0\), we equate equation (1) and equation (2): \[ \lim_{x \to 0} f(x) = f(0) \quad \Rightarrow \quad 2 = k \] Thus, the value of the parameter \(k\) must equal $2$, which matches option (D).