Question

If \[ f(x)= \begin{cases} \dfrac{\sin x}{x}+\cos x, & x\neq 0,\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then the value of \(k\) is: 

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

Hint:

Continuity at a point means "Limit = Value".
The limit \(\lim_{x \to 0} \frac{\sin x}{x}\) is a fundamental result in calculus; ensure it is memorized.
Check individual limits of terms if the overall expression looks complex.

Answer 1

The Correct Option is D

Concept:
• A function \(f(x)\) is continuous at \(x = a\) if \(\lim_{x \to a} f(x) = f(a)\).
• Standard limit: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
• Standard limit: \(\lim_{x \to 0} \cos x = 1\).

Step 1: Calculate the limit of the function as \(x\) approaches 0
To find \(\lim_{x \to 0} f(x)\), we use the expression for \(x \neq 0\):
\[ L = \lim_{x \to 0} \left( \frac{\sin x}{x} + \cos x \right) \]

Step 2: Apply the sum rule of limits
The limit of a sum is the sum of the limits:
\[ L = \lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} \cos x \] Substituting the standard limits:
\[ L = 1 + 1 = 2 \]

Step 3: Equate the limit to the functional value at \(x = 0\)
For continuity at \(x = 0\), we must have:
\[ \lim_{x \to 0} f(x) = f(0) \] From the function definition, \(f(0) = k\).
From our calculation, \(\lim_{x \to 0} f(x) = 2\).
Therefore, \(k = 2\).