Question

If \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} \frac{x - \lfloor x \rfloor}{x - 2} & \text{if } x = 2
\frac{|x - \lfloor x \rfloor|}{a^2 + (x - \lfloor x \rfloor)^2} & \text{if } 1<x<2
2a - b & \text{if } x = 1 \end{cases} \] Then the limit \( \lim_{x \to 0} \frac{\sin(ax) + x \tan(bx){x^2} \) is:}

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

When dealing with limits of trigonometric functions, use standard trigonometric limits like \( \lim_{x \to 0} \frac{\sin(kx)}{x} = k \).

Answer 1

The Correct Option is C

We are given the function \( f(x) \) defined piecewise. We need to find the value of the limit: \[ \lim_{x \to 0} \frac{\sin(ax) + x \tan(bx)}{x^2} \] Step 1: Apply standard limits for trigonometric functions Using the standard limits: \[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \quad \text{and} \quad \lim_{x \to 0} \frac{\tan(kx)}{x} = k \] we can rewrite the expression for small \( x \) as: \[ \frac{\sin(ax)}{x} + \frac{x \tan(bx)}{x^2} \] Step 2: Simplify the limit Now, we simplify each term: \[ \frac{\sin(ax)}{x} = a \quad \text{and} \quad \frac{x \tan(bx)}{x^2} = b \] Thus, the limit becomes: \[ a + b \] Given that \( a = 2 \) and \( b = 0 \), the value of the limit is: \[ 2 + 0 = 2 \] Thus, the correct answer is option (3), \( 2 \).