Let L = lim_{x rightarrow 0} frac{a - sqrt{a^ | vedclass

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

Question

(B) We have $L = \lim_{x \rightarrow 0} \frac{a - a(1 - \frac{x^2}{a^2})^{1/2} - \frac{x^2}{4}}{x^4}$.Using the binomial expansion $(1 - u)^{1/2} = 1

Vedclass · Accepted Answer

$(A, C)$

Vedclass · Answer

(B) We have $L = \lim_{x ightarrow 0} \frac{a - a(1 - \frac{x^2}{a^2})^{1/2} - \frac{x^2}{4}}{x^4}$.Using the binomial expansion $(1 - u)^{1/2} = 1 - \frac{1}{2}u - \frac{1}{8}u^2 - \dots$ where $u = \frac{x^2}{a^2}$:$L = \lim_{x}$ ${ightarrow 0} \frac{a - a(1 - \frac{1}{2}(\frac{x^2}{a^2}) - \frac{1}{8}(\frac{x^2}{a^2})^2) - \frac{x^2}{4}}{x^4}$$L = \lim_{x ightarrow 0} \frac{a - a + \frac{x^2}{2a} + \frac{x^4}{8a^3} - \frac{x^2}{4}}{x^4}$$L = \lim_{x ightarrow 0} \frac{x^2(\frac{1}{2a} - \frac{1}{4}) + \frac{x^4}{8a^3}}{x^4}$.For the limit to be finite,the coefficient of $x^2$ must be zero:$\frac{1}{2a} - \frac{1}{4} = 0 \implies a = 2$.Substituting $a = 2$ into the expression:$L = \lim_{x ightarrow 0} \frac{\frac{x^4}{8(2)^3}}{x^4} = \frac{1}{8 	imes 8} = \frac{1}{64}$.Thus,$a = 2$ and $L = \frac{1}{64}$.Therefore,options $(A)$ and $(C)$ are correct.

Let $L = \lim_{x \rightarrow 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}$,where $a > 0$. If $L$ is finite,then which of the following is true?

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^n} - 1}}{x} = $

The value of $f(0)$ so that the function $f(x) = \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is

$\mathop {\lim }\limits_{n \to \infty } \sin (\pi \sqrt {{n^2} + 1} ) = $

The value of $\lim _{x \rightarrow 0} \left( \frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \right)$ is equal to:

Evaluate the limit: $\lim _{x \rightarrow \infty}\left\{x-\sqrt[n]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}$, where $a_1, a_2, \ldots, a_n$ are positive rational numbers.

Vedclass Test Series

Exam Paper Generator

Online Exam Module