Let beta = lim_{x to 0} frac{alpha x - (e^{3x | vedclass

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

Question

(C) Given $\beta = \lim_{x 	o 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$.Using the expansion $e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \do

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(C) Given $\beta = \lim_{x 	o 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$.Using the expansion $e^{3x} = 1 + 3x + \frac{(3x)^2}{2!} + \dots$,we have:$\beta = \lim_{x 	o 0} \frac{\alpha x - (1 + 3x + \frac{9x^2}{2} + \dots - 1)}{\alpha x(3x + \frac{9x^2}{2} + \dots)}$$\beta = \lim_{x 	o 0} \frac{(\alpha - 3)x - \frac{9}{2}x^2 - \dots}{3\alpha x^2 + \dots}$.For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $\alpha - 3 = 0$,which gives $\alpha = 3$.Substituting $\alpha = 3$ into the expression:$\beta = \lim_{x 	o 0} \frac{-\frac{9}{2}x^2}{3(3)x^2} = \frac{-9/2}{9} = -\frac{1}{2}$.Finally,$\alpha + \beta = 3 - \frac{1}{2} = \frac{5}{2}$.

Let $\beta = \lim_{x \to 0} \frac{\alpha x - (e^{3x} - 1)}{\alpha x(e^{3x} - 1)}$ for some $\alpha \in R$. Then the value of $\alpha + \beta$ is:

Similar Questions

If $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$,then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$

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:

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - ax + b}}{{x - 1}} = 3$,then $a + b$ is equal to

If $\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5$,then $k$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module