Let $L=\lim_{x\to 0} \frac{ a-\sqrt {a^2-x^2} -\frac{x^2}{2}}{x^4}$, $a>0$. If $L$ is finite, find $a$ and $L$ I need a hint to start this question, because I have no idea how to do it. It’s a $\frac 00$ form, so L’Hospital can be applied, but that would be extremely tedious. Expansion can’t be used because there is no function to use it for. How do I start it?
 A: I don't recommend L'Hospital in general. Here a hint : rewrite the expression as
$$ \frac{ a\Bigl(1-\sqrt {1-\frac{x^2}{a^2}}\Bigr) -\frac{x^2}{2}}{x^4}$$
and use Taylor's expansion of the square root. It has a finite limit at $0$ if and only if the principal part of the expansion of the numerator has degree $\ge 4$. Deduce from this observation the value of $a$, then the principal part of the numerator.
A: Hint: you rewrite your limit as follows
$$L=a {1-\left(1-{x^2\over a^2}\right)^{1\over 2}-{x^2\over 2}\over x^4}$$
and you expand using
$$\begin{align}
(1-x)^\alpha=1-&\alpha x\\
+&{\alpha(\alpha-1)\over 2}x^2+o(x^2)
\end{align}$$
and you replace x in the expansion by $x^2/a^2$
A: Idea :
$\displaystyle a-\sqrt{a^2-x^2}-\frac{x^2}{2}=\frac{x^2}{a+\sqrt{a^2-x^2}}-\frac{x^2}{2}$
So the limit to calculate is $\displaystyle\frac{\displaystyle \frac{1}{a+\sqrt{a^2-x^2}}-\frac 1 2}{x^2}$.
To "counter" the effects of the denominator, what should the numerator be equal to?
A: Hint:
$$
\sqrt {a^2  - x^2 }  = a\sqrt {1 - \frac{{x^2 }}{{a^2 }}}  = a - \frac{{x^2 }}{{2a}} - \frac{{x^4 }}{{8a^3 }} +  \cdots ,
$$
when $|x|<a$.
A: You have
$$\frac{a - \sqrt{a^2-x^2}- \frac{x^2}{2}}{x^4} = \frac{a-a\sqrt{1-\frac{x^2}{a^2}} - \frac{x^2}{2}}{x^4} = \frac{a-a\left( 1- \frac{x^2}{2a^2}-\frac{x^4}{8a^4} + o(x^5)\right) - \frac{x^2}{2}}{x^4}$$
$$= \frac{\frac{x^2}{2} \left( \frac{1}{a}-1\right)+\frac{x^4}{8a^3} + o(x^5) }{x^4}$$
You see that if $a \neq 1$, the limit is infinite.
For the limite to be finite, you must have $a=1$, and then you have $$\frac{1 - \sqrt{1^2-x^2}- \frac{x^2}{2}}{x^4} = \frac{1}{8} + o(x)$$
which tends to $\frac{1}{8}$.
