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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 $\sqrt{1-0} = \sqrt{1} = 1$.


$\lim_{x \to \infty} [x (1-1)] = \lim_{x \to \infty} [x*0] = 0$

Where did I go wrong?

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You multiply with $x$. You have $\lim\limits_{x\to\infty} \frac{a}{x} = 0$, for any $a$. But obviously not $\lim\limits_{x\to\infty} \left(x\cdot\frac{a}{x}\right)$. –  Daniel Fischer Oct 29 '13 at 13:53
$\infty .0 $ is an indeterminate form so you can not say $\lim_{x \to \infty} [x.0]=\infty .0=0 $. –  Ömer Oct 29 '13 at 13:59
The error in your approach stems from the common belief that we can replace small parts of a big expression with their limits as and when it suits us. This is not really true. We can replace small parts of a big expression with their limits only when it is allowed via the usual rules of limits with regards to common operations like (add/sub/mul/div) used to various expressions. In most cases such replacements lead to issues like this question and we should avoid this unless we really know the rules of limits clearly. –  Paramanand Singh Oct 30 '13 at 4:44
and by the way as most people are objecting to your handling of $\lim_{x \to \infty}(x \cdot 0) = 0$, I have to disagree here. We do have $\lim_{x \to \infty}(x \cdot 0) = 0$ and there is nothing wrong in this step. The problem happened when you replaced $c/x$ with $\lim_{x \to \infty} c/x = 0$. –  Paramanand Singh Oct 30 '13 at 4:48

3 Answers 3

up vote 1 down vote accepted

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.

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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.

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1) put $ x=\frac{1}{y}$ and consider the limit as $y$ goes to zero.

2) use L'hopital's rule.

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