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The original problem is to calculate $\lim_{x \to 0}\dfrac{1-\sqrt{1-x^2}}{x}$
I simplified the expression to $\lim_{x\to 0}\dfrac{x}{1 + \sqrt{1-x^2}}$
The only definitions and theorems I can use are the definition of a limit and the theorems which states that for two functions $f$ and $g$ that approaches $L_1$ and $L_2$, respectively, near $a$ it is true that
(1.) $\lim_{x\to a} f + g = L1 + L2$
(2.) $\lim_{x\to a} fg = L_1L_2$
(3.) $\lim_{x\to a} \dfrac{1}{f} = \dfrac{1}{L_1}, \quad $ if $L_1 \neq 0$

In order to use (2.) for the simplified expression I first need to establish that I can use (3.) by showing that $\lim_{x\to 0} 1 + \sqrt{1-x^2} \neq 0, \quad$ so I need to find $\lim_{x\to 0} \sqrt{1-x^2}$ with the help of the definition, since none of the theorems says anything about the composition of functions. I know intuitively that the limit is $1$, so I tried to work out a suitable epsilon-delta proof, but I am stuck, because I feel like requiring $|x| < \delta$ will only make $|1 + \sqrt{1-x^2} - 1| = \sqrt{1-x^2}$ bigger than some $\epsilon$, not smaller.

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  • $\begingroup$ @SimpleArt What do you mean? I can't just plug in some number. I need to use the definition or the theorems which I listed in my post because the author has told me of nothing else. $\endgroup$ May 2, 2016 at 21:57
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    $\begingroup$ You need to consider $|1+\sqrt{1-x^2}-\color{red}{2}|$ instead, showing that $2$ is the limit of your expression. Alternatively, consider $|\sqrt{1-x^2}-1|$. You seem to use a mix of these two, which results in your current issues. $\endgroup$
    – Arthur
    May 2, 2016 at 21:57
  • $\begingroup$ @hampadampadoo At $\lim_{x\to0}\sqrt{1-x^2}$, just straight plug in $x=0$, no? $\endgroup$ May 2, 2016 at 21:58
  • $\begingroup$ After your reduction, note that $1+\sqrt{1-x^2}\ge 1$, so the absolute value of $f(x)-0$ is $\le |x|$. $\endgroup$ May 2, 2016 at 22:00
  • $\begingroup$ @SimpleArt And how is this using the definition of a limit? How does it prove anything? $\endgroup$ May 2, 2016 at 22:01

2 Answers 2

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You want to see that the solutions of the inequality $|\sqrt{1-x^2}-1|<\varepsilon$ fill a neighborhood of $0$.

The inequality is equivalent to $$ 1-\varepsilon<\sqrt{1-x^2}<1+\varepsilon $$ and the part $\sqrt{1-x^2}<1+\varepsilon$ holds for every $x$ in the domain of the function. So we need to compute the solutions for $$ 1-\varepsilon<\sqrt{1-x^2} $$ It is not restrictive to assume $0<\varepsilon\le 1$, so the inequality becomes $$ 1-2\varepsilon+\varepsilon^2<1-x^2 $$ that's satisfied for $$ x^2<\varepsilon(2-\varepsilon) $$ so for $$ |x|<\sqrt{\varepsilon(2-\varepsilon)}=\delta $$

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Let $f(x)$ be our function. We want to show that for any given $\epsilon\gt 0$, there is a $\delta$ such that if $0\lt |x-0|\lt\delta$, then $|f(x)-0|\lt \epsilon$.

Note that $1+\sqrt{1-x^2}\ge 1$, at least when $|x|\le 1$. (When $|x|\gt 1$, it is not defined.) It follows that for such $x$ we have $$\left|\frac{x}{1+\sqrt{1-x^2}}\right|\le |x|.$$

Let $\delta=\min(1,\epsilon)$. If $0\lt |x-0|\lt \delta$, then $|f(x)-0|\lt \epsilon$.

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