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

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(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

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$(A, C)$

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(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?

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