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 $L$

Question

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?

Answer 1

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?

Answer 2

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 $|x|<a$.

Answer 3

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 $\frac{1}{8}$.

Answer 4

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.

Answer 5

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 $x^2/a^2$