How to show $\sec^{-1}(1+x) = O(x^{1/2})$ as $x\rightarrow 0$ How to show the following: 
$$\sec^{-1}(1+x) = O(x^{1/2})$$ as $x\rightarrow 0$  
According to the definition, I have to find the positive constant $C$, i.e., I want to obtain:
$$\lim_{x\rightarrow 0} |\frac{\sec^{-1}(1+x)}{x^{1/2}}| = C$$
Now use the L'hospital's rule:  
$$\lim_{x\rightarrow 0} |\frac{\sec^{-1}(1+x)}{x^{1/2}}| = \lim_{x\rightarrow 0} |\frac{\frac{1}{(1+x) \sqrt{(1+x)^2-1}}}{\frac{1}{2}x^{-1/2}}| = \lim_{x\rightarrow 0} |\frac{2x^{1/2}}{(1+x) \sqrt{(1+x)^2-1}}|  $$ 
I have no idea how to do the next step. The numerator is always $0$! How to find a positive constant $C$ to show this? 
Please advise !
 A: $$\lim_{x\rightarrow 0} |\frac{2x^{1/2}}{(1+x) \sqrt{(1+x)^2-1}}|=\lim_{x\rightarrow 0} |\frac{2x^{1/2}}{(1+x) \sqrt{x^2+2x}}|=\lim_{x\rightarrow 0} |\frac{2x^{1/2}}{(1+x) {\sqrt{x}\sqrt{x+2}}}|=lim_{x\to 0}\frac{2}{\sqrt{x+2}}=\sqrt 2$$
A: You should consider domain and range of $\sec^{-1}(x)$. Lets say the function is $\Bbb R \setminus (1,1) \to [0,\pi] \setminus \{\pi / 2\}$. 
Then we can say that for function $\sec^{-1}(1+x)$, $\Bbb R \setminus (-2,0) \to [0,\pi] \setminus \{\pi / 2\}$.
So evident from here, the left hand limit is not existent. Only we can talk about right hand limit. This is also evident since $\sec(x)$ never lies in $(-1,1)$.
Another approach is to put $\sec^{-1}(1+x) = t$ so $\sec(t) = 1+x$. Note as $x\to 0, t\to 0$. Then $\cos(t) = \frac{1}{1+x}=1-x+x^2 ...$
Now using $\cos(t) \approx 1- \frac{t^2}{2}$ and $\frac{1}{1+x} \approx 1-x$ (neglecting higher power terms as $t,x \to 0$), we can say $1-\frac{t^2}{2} \approx 1-x$ or $t \approx \sqrt{2x}$.
Note this is only valid for $x\to 0^+$ as earlier said!
