Step 1: Understanding the Concept:
For a limit to exist finitely when the denominator goes to 0, the numerator must also go to 0.
As \( x \to 2 \), denominator \( \to (\sqrt{1}-1)\log(1) = 0 \times 0 = 0 \). Step 2: Key Formula or Approach:
Numerator \( \sin(x^3 - 5x^2 + ax + b) \to 0 \) at \( x=2 \).
This implies \( 2^3 - 5(2^2) + a(2) + b = 0 \implies 2a + b = 12 \) \dots (1). Step 3: Detailed Explanation:
Let \( x - 2 = h \), as \( x \to 2 \), \( h \to 0 \).
Denominator \( = (\sqrt{h+1}-1)\log(1+h) \approx \frac{h}{2} \cdot h = \frac{h^2}{2} \).
Since the denominator is \( O(h^2) \), the numerator must also be at least \( O(h^2) \).
Let \( f(x) = x^3 - 5x^2 + ax + b \). Since \( f(2)=0 \) and it must contain \( (x-2)^2 \) for the limit to be finite:
\( f'(2) = 0 \implies 3x^2 - 10x + a = 0 \) at \( x=2 \).
\( 3(4) - 20 + a = 0 \implies a = 8 \).
From (1), \( 2(8) + b = 12 \implies b = -4 \).
Now the numerator is \( \sin((x-2)^2(x-1)) \).
Limit \( m = \lim_{h \to 0} \frac{\sin(h^2(1))}{\frac{h^2}{2}} = 2 \).
Value \( a + b + m = 8 - 4 + 2 = 6 \). Step 4: Final Answer:
The final value of \( a + b + m \) is 6.
