Question

If \(a\) and \(b\) are non-negative real numbers and \[ \lim_{x\to0}\frac{e^{ax}-\cos bx}{1-\cos x}=4, \] then \[ \lim_{x\to a}\frac{\sin(bx-ab)}{x-a} = \]

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(8\)

Hint:

Remember the standard limits \[ \boxed{ \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12, \qquad \lim_{x\to0}\frac{\sin x}{x}=1. } \] Also, if a limit is finite, first eliminate any lower-order terms in the numerator.

Answer 1

The Correct Option is B

Step 1: Evaluate the given limit. Using the expansions near \(x=0\), \[ e^{ax}=1+ax+\frac{a^2x^2}{2}+O(x^3), \] and \[ \cos bx=1-\frac{b^2x^2}{2}+O(x^4). \] Hence, \[ e^{ax}-\cos bx = ax+\frac{a^2+b^2}{2}x^2+O(x^3). \] Also, \[ 1-\cos x=\frac{x^2}{2}+O(x^4). \] Since the given limit is finite, \[ a=0. \] Therefore, \[ \lim_{x\to0} \frac{\dfrac{b^2x^2}{2}}{\dfrac{x^2}{2}} = b^2. \] Given that the limit equals \(4\), \[ b^2=4. \] Since \(b\ge0\), \[ \boxed{b=2.} \]

Step 2: Evaluate the required limit. Since \[ a=0, \] the required limit becomes \[ \lim_{x\to0}\frac{\sin(2x)}{x}. \] Using the standard limit, \[ \lim_{t\to0}\frac{\sin t}{t}=1, \] we obtain \[ \lim_{x\to0}\frac{\sin(2x)}{x} = 2. \] Hence, \[ \boxed{ \lim_{x\to a}\frac{\sin(bx-ab)}{x-a}=2. } \] Therefore, the correct option is \(\boxed{(B)}\).