Derivative of secant using derivative definition Is there any to get
$$\lim_{h\to 0} \frac{\sec{(x+h)} - \sec{x}}{h} = \sec x \tan x.$$
What I mean is, proving the derivative of $\sec{x}$, without using product/quotient rule or the sine/cosine derivatives, just limits. Pure limits.
EDIT: I'm sorry, i'm just learning how to write formulas correctly.
 A: Hint:
$$\frac{\sec(x+h)-\sec x}{h}= \frac{\cos x-\cos(x+h)}{h}\frac{1}{\cos x\cos(x+h)} \\=\frac{1 - \cos h}{h}\frac{1}{\cos(x+h)}+ \frac{\sin h}{h}\frac{\sin x}{\cos x \cos(x+h)}$$
A: $$\begin{array}{lll}
\displaystyle\lim_{h\to 0}\frac{\sec(x+h)-\sec x}{h} &=& \displaystyle\lim_{h\to 0}\frac{\frac{\sec h\sec x}{1-\tan h \tan x}-\sec x}{h}\\
&=& \displaystyle\lim_{h\to 0}\frac{\sec h\sec x-\sec x+\tan h\sec x\tan x}{h(1-\tan h \tan x)}\\
&=& \displaystyle\lim_{h\to 0}\frac{\sec h\sec x-\sec x}{h(1-\tan h \tan x)}+\displaystyle\lim_{h\to 0}\frac{\tan h\sec x\tan x}{h(1-\tan h \tan x)}\\
&=& \displaystyle\sec x\lim_{h\to 0}\frac{\sec h}{1-\tan h\tan x}\cdot\color{green}{\frac{\sec h-1}{h\sec x}}+\displaystyle\lim_{h\to 0}\frac{\tan h\sec x\tan x}{h(1-\tan h \tan x)}\\
&=& \displaystyle\sec x\cdot 1\cdot0+\displaystyle\lim_{h\to 0}\frac{\tan h\sec x\tan x}{h(1-\tan h \tan x)}\\
&=& \displaystyle\lim_{h\to 0}\frac{\tan h\sec x\tan x}{h(1-\tan h \tan x)}\\
&=& \displaystyle\lim_{h\to 0}\color{green}{\frac{\tan h}{h\sec h}}\cdot\frac{\sec h}{(1-\tan h \tan x)}\cdot\sec x \tan x\\
&=& \displaystyle\lim_{h\to 0}1\cdot1\cdot\sec x \tan x\\
&=& \displaystyle\sec x \tan x\\
\end{array}$$
Now all that needs to be shown is that
$$\lim_{h\to 0}\frac{\tan h}{h\sec h} = 1\text{ and }\lim_{h\to 0}\frac{\sec h - 1}{h\sec h} = 0$$
I'll let you do that.
Hint: a circle sector with an angle of $h$, and a radius of $1$ is similar to a circle sector with an angle of $h$ and a radius of $\sec h$ [draw a diagram].
