<s>[INST] Suppose $\lim _{ x\to 0 } \frac { a-\sqrt { { a }^{ 2 }-{ x }^{ 2 } }-{ x }^{ 2 }/4 }{ { x }^{ 4 } } $ is finite then how to find the value of a? Suppose $\lim _{ x\to 0 } \frac { a-\sqrt { { a }^{ 2 }-{ x }^{ 2 } }-{ x }^{ 2 }/4  }{ { x }^{ 4 } } $ is finite then how to find the value of a ?
I'm having trouble understand the fact that even though the highest power of x in the denominator and numerator is not same,how can the limit be finite?
Yeah I know,probably i'm having a conceptual doubt.Can someone help please!
 [/INST] Answering the question about why these may vanish to the same order: 
First, since $x$ is approaching $0$, you should be looking at the lowest order terms (the highest order terms are used when $x$ approaches infinity).
The numerator can be broken into 
$$
\left(a-\frac{x^2}{4}\right)-\sqrt{a^2-x^2}
$$
The lowest order term of the first part is $a$ and the lowest order term of the second term is also $a=\sqrt{a^2}$.  Therefore, these two lowest order terms cancel and the vanishing is of higher order.
For a more hands-on example, consider $x+x^2$ and $x-x^3$.  At zero, both of these vanish to first order because of the $x$'s, but their difference vanishes to second order because the difference is $x^2+x^3$.
To actually solve this problem, you can take the original limit:
$$
\lim_{x\rightarrow 0}\frac{a-\frac{x^2}{4}-\sqrt{a^2-x^2}}{x^4}
$$
and multiply the numerator and denominator by $a-\frac{x^2}{4}+\sqrt{a^2-x^2}$, the conjugate of the numerator.  This gives
$$
\lim_{x\rightarrow 0}\frac{\left(a-\frac{x^2}{4}\right)^2-(a^2-x^2)}{x^4\left(a-\frac{x^2}{4}+\sqrt{a^2-x^2}\right)}
$$
Expanding out (and simplifying the numerator) we get
$$
\lim_{x\rightarrow 0}\frac{-\frac{ax^2}{2}+\frac{x^4}{16}+x^2}{x^4\left(a-\frac{x^2}{4}+\sqrt{a^2-x^2}\right)}
$$
Since $x$ is approaching zero, the smallest terms in the numerator and denominator matter the most.  If $(1-a/2)\not=0$, then the numerator has a $x^2$ term and the denominator has an $x^4$ term, which result in $\frac{1}{x^2}$, which diverges near $0$.  Therefore, the $x^2$ term in the numerator must be eliminated, so $a=2$.
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