Question

If $\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1}-1) \log_e(x-1)} = m$ (exists finitely), then find the value of $a + b + m$.

Answer 1

Step 1: Analyze the limit existence.
As $x \to 2$, the denominator $(\sqrt{2-1}-1) \log_e(2-1) = (1-1) \log(1) = 0 \cdot 0 = 0$.
For the limit to exist finitely, the numerator must also be 0 at $x=2$.
$\sin(2^3 - 5(2^2) + a(2) + b) = 0 \implies 8 - 20 + 2a + b = 0 \implies 2a + b = 12$ ---(1).


Step 2: Simplify the limit using standard limits.
Divide and multiply the numerator by its argument $u = x^3 - 5x^2 + ax + b$.
$\lim_{x \to 2} \frac{\sin(u)}{u} \cdot \frac{u}{(\sqrt{x-1}-1) \log_e(1 + (x-2))}$
Since $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{h \to 0} \frac{\log(1+h)}{h} = 1$ (where $h = x-2$):
The limit becomes $\lim_{x \to 2} \frac{u}{(\sqrt{x-1}-1) (x-2)}$.
Rationalize the square root term: $\sqrt{x-1}-1 = \frac{(x-1)-1}{\sqrt{x-1}+1} = \frac{x-2}{\sqrt{x-1}+1}$.
Limit $= \lim_{x \to 2} \frac{u}{\frac{x-2}{\sqrt{x-1}+1} \cdot (x-2)} = \lim_{x \to 2} \frac{u(\sqrt{x-1}+1)}{(x-2)^2}$
As $x \to 2$, $\sqrt{x-1}+1 \to 2$.
So $m = 2 \lim_{x \to 2} \frac{x^3 - 5x^2 + ax + b}{(x-2)^2}$.


Step 3: Apply L'Hopital's rule.
Since the denominator is 0 at $x=2$, the numerator must be 0 (already established $2a+b=12$).
$m = 2 \lim_{x \to 2} \frac{3x^2 - 10x + a}{2(x-2)}$.
Again, for this to exist, numerator must be 0 at $x=2$:
$3(4) - 10(2) + a = 0 \implies 12 - 20 + a = 0 \implies a = 8$.
Substitute $a=8$ into (1): $2(8) + b = 12 \implies b = -4$.


Step 4: Find $m$.
$m = \lim_{x \to 2} \frac{3x^2 - 10x + 8}{x-2} = \lim_{x \to 2} \frac{(3x-4)(x-2)}{x-2} = 3(2) - 4 = 2$.


Step 5: Calculate $a+b+m$.
$a + b + m = 8 + (-4) + 2 = 6$.
The answer is 6.