Show that $\lim_{x \to 0}\sqrt{1-x^2} = 1$ with the help of the definition of a limit 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.
 A: 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
$$
|x|<\sqrt{\varepsilon(2-\varepsilon)}=\delta
$$
A: 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 $|f(x)-0|\lt \epsilon$. 
