If beta = lim_{x rightarrow 0} frac{e^{x^3} - | vedclass

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

Question

(A) We are given $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.Using the standard limit

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) We are given $\beta = \lim_{x ightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$.Using the standard limit $\lim_{x ightarrow 0} \frac{\sin x}{x} = 1$,we can rewrite the denominator as $x \sin^2 x = x^3 \left(\frac{\sin x}{x}ight)^2 \approx x^3$ as $x ightarrow 0$.Expanding the terms using Taylor series:$e^{x^3} = 1 + x^3 + O(x^6)$$(1 - x^3)^{1/3} = 1 - \frac{1}{3}x^3 + O(x^6)$$(1 - x^2)^{1/2} - 1 = (1 - \frac{1}{2}x^2 + O(x^4)) - 1 = -\frac{1}{2}x^2 + O(x^4)$Substituting these into the expression:$\beta = \lim_{x ightarrow 0} \frac{(1 + x^3) - (1 - \frac{1}{3}x^3) + (-\frac{1}{2}x^2) \cdot x}{x^3}$$\beta = \lim_{x ightarrow 0} \frac{1 + x^3 - 1 + \frac{1}{3}x^3 - \frac{1}{2}x^3}{x^3}$$\beta = \lim_{x}$ ${ightarrow 0} \frac{(1 + \frac{1}{3} - \frac{1}{2})x^3}{x^3} = 1 + \frac{1}{3} - \frac{1}{2} = \frac{6 + 2 - 3}{6} = \frac{5}{6}$.Therefore,$6 \beta = 6 	imes \frac{5}{6} = 5$.

If $\beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$,then the value of $6 \beta$ is

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4}$

If $f(x) = \begin{cases} x & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$,then $\mathop {\lim }\limits_{x \to 0} f(x) = $

$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

Let $f(x) = \frac{\sqrt{x+3}}{x+1}$. Then the value of $\lim_{x \rightarrow -3^{-}} f(x)$ is

$\lim \limits_{x}$ ${\rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\cos ^{-1} x\right)-x}{1-\tan \left(\cos ^{-1} x\right)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module