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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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:

Similar Questions

If $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+a^2\right)\right\}^{1 / x}=2 a \sin ^2 \theta$, where $a>0$ and $\theta \in R$, then:

If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ is:

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$, where $a, b \in R$, then:

Let $a$ be an integer such that $\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x]-3a}$ exists,where $[t]$ denotes the greatest integer function $\leq t$. Then $a$ is equal to:

If $\lim_{x \rightarrow 0} \frac{e^{(a-1)x} + 2 \cos(bx) + (c-2)e^{-x}}{x \cos x - \log_{e}(1+x)} = 2$, then $a^{2} + b^{2} + c^{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module