Question

If \[ \lim_{x\to 2}\frac{\sin(x^3-5x^2+ax+b)}{(\sqrt{x-1}-1)\log_e(x-1)}=m, \] then \(a+b+m\) is equal to:

• A. \(5\)
• B. \(6\)
• C. \(8\)
• D. \(10\)

Answer 1

The Correct Option is C

Concept: When evaluating limits involving \(\sin f(x)\), we use the standard result: \[ \lim_{x\to0}\frac{\sin x}{x}=1 \] Thus the expression inside sine must approach \(0\).
Step 1:Ensure the numerator tends to zero.} \[ x^3-5x^2+ax+b \] At \(x=2\): \[ 8-20+2a+b=0 \] \[ 2a+b=12 \]
Step 2:Approximate the denominator.} Near \(x=2\), \[ \sqrt{x-1}-1 \approx \frac{x-2}{2} \] \[ \log(x-1) \approx x-2 \] Thus denominator behaves like \[ \frac{x-2}{2}(x-2)=\frac{(x-2)^2}{2} \]
Step 3:Expand the numerator near \(x=2\).} \[ f(x)=x^3-5x^2+ax+b \] \[ f'(x)=3x^2-10x+a \] \[ f'(2)=12-20+a=a-8 \] Hence \[ f(x)\approx (a-8)(x-2) \] Thus \[ \sin f(x)\approx (a-8)(x-2) \]
Step 4:Compute the limit.} \[ m=\lim_{x\to2}\frac{(a-8)(x-2)}{\frac{(x-2)^2}{2}} \] \[ m=\frac{2(a-8)}{x-2} \] For finite limit we require \[ a=8 \] Thus from \[ 2a+b=12 \] \[ 16+b=12 \] \[ b=-4 \] Then \[ m=4 \]
Step 5:Find the required value.} \[ a+b+m=8-4+4 \] \[ =8 \]