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

Define $f(x) = \begin{cases} b - ax & \text{if } x < 2 \\ 3 & \text{if } x = 2 \\ a + 2bx & \text{if } x > 2 \end{cases}$. If $\lim_{x \rightarrow 2} f(x)$ exists,then find the value of $\frac{a}{b}$.

The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

Let $\tan (2\pi |\sin \theta |) = \cot (2\pi |\cos \theta |)$,where $\theta \in R$ and $f(x) = (\sin^2 \theta + \cos^2 \theta)$. The value of $\lim_{x \to \infty} [\frac{2}{f(x)}]$ equals (Here $[\,]$ represents the greatest integer function).

If $a$ and $b$ are roots of the equation $px^2 + qx + r = 0$,then $\lim_{x \rightarrow b} \frac{1 - \cos 2(px^2 + qx + r)}{2(px - pb)^2}$ is equal to

If $\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to 0} \frac{1 - \cos(2x)}{x \sin x}$, then the value of $k$ is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module