For alpha, beta, gamma in R,if lim _{x righta | vedclass

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

Question

(A) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}}{\sin(2x) - \beta x} = 3$. For the limit to exist and be f

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Given the limit: $\lim _{x \rightarrow 0} \frac{x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}}{\sin(2x) - \beta x} = 3$. For the limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$. Since $\lim _{x \rightarrow 0} (x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}) = 0 + (\gamma-1)(1) = \gamma-1$,we must have $\gamma-1 = 0$,so $\gamma = 1$. Now the expression becomes $\lim _{x \rightarrow 0} \frac{x^2(\alpha x + O(x^3)) + 0(e^{x^2}-1)}{(2x - \frac{8x^3}{6} + O(x^5)) - \beta x} = \lim _{x \rightarrow 0} \frac{\alpha x^3}{(2-\beta)x - \frac{4}{3}x^3} = 3$. For the limit to be non-zero,the denominator must have the same power of $x$ as the numerator. Thus,the coefficient of $x$ must be $0$,so $2-\beta = 0$,which gives $\beta = 2$. Now the limit is $\lim _{x \rightarrow 0} \frac{\alpha x^3}{-\frac{4}{3}x^3} = -\frac{3\alpha}{4} = 3$. Solving for $\alpha$,we get $\alpha = -4$. Finally,$\beta + \gamma - \alpha = 2 + 1 - (-4) = 7$.

For $\alpha, \beta, \gamma \in R$,if $\lim _{x \rightarrow 0} \frac{x^2 \sin(\alpha x) + (\gamma-1) e^{x^2}}{\sin(2x) - \beta x} = 3$,then $\beta + \gamma - \alpha$ is equal to:

Similar Questions

Let $a > 0$ be a root of the equation $2x^2 + x - 2 = 0$. If $\lim_{x \rightarrow \frac{1}{a}} \frac{16(1 - \cos(2 + x - 2x^2))}{1 - ax^2} = \alpha + \beta \sqrt{17}$,where $\alpha, \beta \in \mathbb{Z}$,then $\alpha + \beta$ is equal to:

If $\mathop {Lim}\limits_{x \to 0} (x^{-3} \sin 3x + ax^{-2} + b)$ exists and is equal to zero,then:

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:

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+x+1}{x+1}-a x-b\right)=4$,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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module