Question on limits

Question

Let $$L=\lim_{x\to \infty} \frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$$.where L is a finite real number. Then value of a,L for which this is true . Options were there answer is $a=2,L=1/64$. $$\text{My Try}$$. I separated first terms of a and x ie first two terms and third term.Thinking quite a while I multiplied by conjugate of the first two terms. Last term is in $x$ so $x^2$ cancels from numerator,denominator so we are left with $1/4x^2$ as x goes to infinity this goes to $0$. So this term is gone. Now I am thinking whether to solve this equation of two unknowns ie $L=\lim_{x\to \infty}\frac{1}{x^2(a+\sqrt{a^2-x^2})}$. Or is there any better approach.

Answer 1

The limit at $\infty$ doesn't make sense, because the function is only defined in $[-|a|,|a|]\setminus\{0\}$.

You probably want \begin{align} \lim_{x\to0}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4} &= \lim_{x\to0}\frac{(a-\frac{x^2}{4})^2-(a^2-x^2)}{x^4} \frac{1}{a-\frac{x^2}{4}+\sqrt{a^2-x^2}} \\ &= \lim_{x\to0}\frac{-\frac{ax^2}{2}+\frac{x^4}{16}+x^2}{x^4} \frac{1}{a-\frac{x^2}{4}+\sqrt{a^2-x^2}} \end{align} and now it should be clear that if $a\ne2$, the limit is not finite.

For $a=2$ the limit is indeed $1/64$.