Question regarding simple limit Why is it, that:
$\lim_{x \to \infty} [x (1-\sqrt{1-\frac{c}{x}})] = \frac{c}{2}$
Link: Wolframalpha
and not $0$?
My (obviously incorrect) reasoning:
Since $c$ is an arbitrary constant, and as $x$ goes to infinity $\frac{c}{x}$ will practically equal $0$. thus $\sqrt{1-0} = \sqrt{1} = 1$.
Therefore, 
$\lim_{x \to \infty} [x (1-1)] = \lim_{x \to \infty} [x*0] = 0$
Where did I go wrong?
 A: Use the fact that 
$$
\sqrt{1 - \frac{c}{x}} = 1 - \frac{c}{2x} + \mathcal{O}\Big (\frac{1}{x^2} \Big )
$$
for $|x|$ large, or multiply numerator and denominator by the "conjugate" 
$$
1 + \sqrt{1 - \frac{c}{x}}, 
$$
and then take the limit. 
A: The reason your approach doesn't work is that you are multiplying $0$ by infinity, which does not have a well defined result. I would probably first substitute $a = \frac{1}{x}$, which gives the equivalent but (in my opinion) simpler
$\lim_{a\downarrow 0} \frac{1}{a}\left( 1 - \sqrt{1 - ca}\right)$
Then substituting the Taylor expansion $\sqrt{1-ca} = 1 - \frac{c}{2}a - \frac{c^2}{8}a^2 + O(a^3)$ one finds
$\lim_{a\downarrow 0} \frac{1}{a}\left(\frac{c}{2}a + \frac{c^2}{8}a^2 + O(a^3)\right) = \lim_{a\downarrow 0} \left(\frac{c}{2} + \frac{1}{8}c^2a + O(a^2)\right) = \frac{c}{2}$
The main thing to keep in mind is that you always want to avoid expressions looking like $0\cdot\infty$ because their value is undefined.
A: Hint:
1) put $ x=\frac{1}{y}$ and consider the limit as $y$ goes to zero.
2) use L'hopital's rule.
