Calculation of $\lim\limits_{x \to 0} \frac{\frac{\mathrm d}{\mathrm d x} (e^{\sec x})}{\frac{\mathrm d}{\mathrm d x} (e^{\sec 2x})}$ I'm a bit rusty with limits and derivatives at the moment. I was doing L'hosp on another problem when I got stuck here.
$$\lim_{x \to 0} \dfrac{\dfrac{\mathrm d}{\mathrm d x} (e^{\large \sec x})}{\dfrac{\mathrm d}{\mathrm d x} (e^{\large\sec 2x})}$$
Further L'hosp is a mess. Care to continue, my jolly fellows?
 A: Let $f(x) = e^{\sec x}$. Then you are looking for 
$$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{\mathrm d}{\mathrm d x} (e^{\sec x})}{\dfrac{\mathrm d}{\mathrm d x} (e^{\sec 2x})} = \lim_{x \to 0}{f'(x) \over 2 f'(2x)}$$
Note that $f'(x) = \tan x \sec x \,e^{\sec x}$, so that $f'(0) = 0$. This means the limit is of $0/0$ form, so we use L'hopital, and the limit becomes
$$= \lim_{x \to 0}{f''(x) \over 4 f''(2x)}$$
Here
 $f''(x) = \sec^3 x \, e^{\sec x} + \tan^2 x \sec x \, e^{\sec x} + (\tan x \sec x)^2\, e^{\sec x}$, and $f''(0) = 1 \neq 0$. Hence the limit is ${1 \over 4}$.
A: Executing the derivatives we get:
$$
\frac{1}{2}\left(\lim_{x\rightarrow 0}\left( \frac{\frac{\sin(x)}{\cos^2(x)}e^{1/\cos(x)}}{\frac{\sin(2x)}{\cos^2(2x)}e^{1/\cos(2x)}} \right)\right)
$$
$=$
$$
\frac{1}{2}\lim_{x\rightarrow 0}\left(\frac{\sin(x)}{\sin(2x)}\right)\left(\lim_{x\rightarrow 0}\left( \frac{\frac{1}{\cos^2(x)}e^{1/\cos(x)}}{\frac{1}{\cos^2(2x)}e^{1/\cos(2x)}} \right)\right)
$$
In the last part we have factored out a surely converging part.
Using that $e^x\approx 1+x$ , $\sin(x)\approx x$ , $\cos(x)\approx 1$ , $e^{1/\cos(ax)}\approx 1+\frac{1}{\cos(ax)}\approx2$ at $x$=0 we can rewrite this as
$$
\frac{1}{2}\left(\lim_{x\rightarrow 0}\left( \frac{\frac{x}{1}2}{\frac{2 x}{1}2} \right)\right)=\frac{1}{4}
$$
