Let L = lim    [a - sqrt(a² - x² ) - (x² /4)] /  x⁴ , a > 0.
            x--0

If L is finite, then

a). a = 2; L = 1/64
b). a = 1; L = 1/64
c). a = 3; L = 1/32
d). a = 1; L = 1/32

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{a}-\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}-{\displaystyle \frac{{\mathrm{x}}^2}{4}}}{{\mathrm{x}}^4} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{\left(\mathrm{a}-{\displaystyle \frac{{\mathrm{x}}^2}{4} }\right)-\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}{{\mathrm{x}}^4}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{\left(\mathrm{a}-{\displaystyle \frac{{\mathrm{x}}^2}{4} }\right)-\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}{{\mathrm{x}}^4}\right\}\left\{\frac{\left(\mathrm{a}-{\displaystyle \frac{{\mathrm{x}}^2}{4} }\right)+\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}{\left(\mathrm{a}-\frac{{\mathrm{x}}^2}{4} \right)+\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\left(\mathrm{a}-{\displaystyle \frac{{\mathrm{x}}^2}{4} }\right)}^2-\left({\mathrm{a}}^2-{\mathrm{x}}^2\right)}{{\mathrm{x}}^4}\right\}\left\{\frac{1}{\left(\mathrm{a}-\frac{{\mathrm{x}}^2}{4} \right)+\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\mathrm{a}}^2+{\displaystyle \frac{{\mathrm{x}}^4}{16}-2\mathrm{a}\frac{{\mathrm{x}}^2}{4} }-{\mathrm{a}}^2+{\mathrm{x}}^2}{{\mathrm{x}}^4}\right\}\left\{\frac{1}{\left(\mathrm{a}-\frac{{\mathrm{x}}^2}{4} \right)+\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle \frac{{\mathrm{x}}^4}{16}-\frac{{\mathrm{ax}}^2}{2} +{\mathrm{x}}^2}}{{\mathrm{x}}^4}\right\}\left\{\frac{1}{\left(\mathrm{a}-0 \right)+\sqrt{{\mathrm{a}}^2-0}}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle \frac{{\mathrm{x}}^4}{16}+{\mathrm{x}}^2\left(1-\frac{\mathrm{a}}{2}\right)}}{{\mathrm{x}}^4}\right\}\left\{\frac{1}{\mathrm{a}+\mathrm{a}}\right\} \\ \mathrm{For} \mathrm{limit} \mathrm{to} \mathrm{be} \mathrm{finite}, {\mathrm{x}}^4 \mathrm{in} \mathrm{denominator} \mathrm{must} \mathrm{get} \mathrm{cancelled}, \mathrm{hence} \mathrm{coefficient} \\ \mathrm{of} {\mathrm{x}}^2 \mathrm{in} \mathrm{numerator} \mathrm{must} \mathrm{be} 0. \\ \Rightarrow 1-\frac{\mathrm{a}}{2}=0 \\ \Rightarrow \mathrm{a}=2 \\ \mathrm{Hence} \\ \mathrm{L}=\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle \frac{{\mathrm{x}}^4}{16}+0}}{{\mathrm{x}}^4}\right\}\left\{\frac{1}{2+2}\right\} \\ =\underset{\mathrm{x}\to 0}{\lim } \left\{\frac{{\displaystyle 1}}{16}\right\}\left\{\frac{1}{4}\right\} \\ =\frac{1}{16\times 4} \\ =\frac{1}{64} \\ \mathrm{Hence} \mathrm{a}=2 \mathrm{and} \mathrm{L}=\frac{1}{64} \end{gathered}$$

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