If lim_{x to 2} frac{sin(x^3 - 5x^2 + ax + b) | vedclass

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(B) For the limit to exist,the numerator must approach $0$ as $x 	o 2$. Thus,$8 - 20 + 2a + b = 0 \Rightarrow 2a + b = 12$.Let $t = x-2$,so $x = t+2$

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) For the limit to exist,the numerator must approach $0$ as $x 	o 2$. Thus,$8 - 20 + 2a + b = 0 \Rightarrow 2a + b = 12$.Let $t = x-2$,so $x = t+2$. As $x 	o 2$,$t 	o 0$.The denominator becomes $(\sqrt{t+1}-1)\log_e(t+1) \approx (t/2)(t) = t^2/2$.The numerator becomes $\sin((t+2)^3 - 5(t+2)^2 + a(t+2) + b) = \sin(t^3 + 6t^2 + 12t + 8 - 5t^2 - 20t - 20 + at + 2a + b) = \sin(t^3 + t^2 + (a-8)t + (2a+b-12))$.Since $2a+b=12$,the expression simplifies to $\sin(t^3 + t^2 + (a-8)t)$.For the limit to exist and be finite,the coefficient of $t$ must be $0$,so $a-8=0 \Rightarrow a=8$.Substituting $a=8$ into $2a+b=12$,we get $16+b=12 \Rightarrow b=-4$.The numerator is now $\sin(t^3+t^2) \approx t^2$ as $t 	o 0$.The limit is $\lim_{t 	o 0} \frac{t^2}{t^2/2} = 2$,so $m=2$.Finally,$a+b+m = 8 - 4 + 2 = 6$.

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:

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