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

Step 1: Understanding the Concept:
For the limit to result in a finite non-zero value \(m\), given that the denominator tends to 0 as \(x \to 2\), the numerator must also tend to 0.
Furthermore, the order of the zero in the denominator must match the order of the zero in the numerator for the limit to be finite.
Step 2: Key Formula or Approach:
1. Use approximations for \(x \to 2\): \(\sin(f(x)) \approx f(x)\), \(\log_e(1+t) \approx t\), and \(\sqrt{1+t}-1 \approx t/2\).
2. If \(x \to a\) is a zero of order \(n\), then \((x-a)^n\) is a factor of the polynomial.
Step 3: Detailed Explanation:
Let \(x - 2 = t\). As \(x \to 2, t \to 0\).
The denominator becomes \((\sqrt{1+t}-1) \log_e(1+t)\).
Applying limits: \(\lim_{t \to 0} \frac{\sqrt{1+t}-1}{t} \cdot t \cdot \frac{\log_e(1+t)}{t} \cdot t = \frac{1}{2} \cdot t^2\).
The denominator is of the order \((x-2)^2\).
Thus, the polynomial \(P(x) = x^3 - 5x^2 + ax + b\) must have \((x-2)^2\) as a factor.
Let \(x^3 - 5x^2 + ax + b = (x-2)^2(x-k)\).
Expanding: \((x^2 - 4x + 4)(x-k) = x^3 - (k+4)x^2 + (4k+4)x - 4k\).
Comparing the coefficients of \(x^2\): \(k+4 = 5 \implies k = 1\).
Now find \(a\) and \(b\):
\(a = 4k + 4 = 4(1) + 4 = 8\).
\(b = -4k = -4(1) = -4\).
Now find \(m\):
\[ m = \lim_{x \to 2} \frac{(x-2)^2(x-1)}{\frac{1}{2}(x-2)^2} = 2 \lim_{x \to 2} (x-1) = 2(2-1) = 2 \]
Calculation of \(a + b + m = 8 + (-4) + 2 = 6\).
Step 4: Final Answer:
The value of \(a+b+m\) is 6.