If _x2(x^3-5x^2+ax+b)(x-1-1)_e(x-1)=m, then a+b+m is equal to:

Question

If $\displaystyle\lim_{x	o2}\frac{\sin(x^3-5x^2+ax+b)}{(\sqrt{x-1}-1)\log_e(x-1)}=m$, then $a+b+m$ is equal to:

Accepted Answer

Let $t=x-2\to0$. As $x\to2$, $\sqrt{x-1}-1=\sqrt{1+t}-1\approx t/2$ and $\log_e(x-1)=\log_e(1+t)\approx t$, so the denominator $\approx t^2/2$.

For the limit to be finite, the cubic $P(x)=x^3-5x^2+ax+b$ must vanish at $x=2$ (so numerator $\to0$):
$$8-20+2a+b=0\implies 2a+b=12 \quad\text{...(1)}$$

Near $x=2$, $\sin P(x)\approx P(x)$. If $P'(2)\ne0$ the numerator would be $O(t)$ against an $O(t^2)$ denominator, making the limit infinite --- so we also need $P'(2)=0$:
$$P'(x)=3x^2-10x+a,\quad P'(2)=12-20+a=a-8=0\implies a=8$$

From (1): $b=12-16=-4$.

With $P'(2)=0$, use the second-order term: $P(x)\approx\dfrac{P''(2)}{2}t^2$, where $P''(x)=6x-10$, $P''(2)=2$, so $P(x)\approx t^2$.

$$m=\lim_{t\to0}\frac{t^2}{t^2/2}=2$$

$a+b+m=8-4+2=6$

Solutions · 1