# Prove that?  lim_(h->0)(sec(x+h) - sec x)/h = sec x*tan x

Nov 16, 2016

See belw

#### Explanation:

We want to prove that

${\lim}_{h \to 0} \frac{\sec \left(x + h\right) - \sec x}{h} = \sec x \cdot \tan x$

Hopefully, you can identify this as the limit used in a derivative, and so this is the same as procing that:

$\frac{d}{\mathrm{dx}} \sec \left(x\right) = \sec x \cdot \tan x$

Let $L = {\lim}_{h \to 0} \frac{\sec \left(x + h\right) - \sec x}{h}$, Then:

$L = {\lim}_{h \to 0} \frac{\frac{1}{\cos \left(x + h\right)} - \frac{1}{\cos x}}{h}$
$\therefore L = {\lim}_{h \to 0} \frac{\cos x - \cos \left(x + h\right)}{h \cos \left(x + h\right) \cos x}$
$\therefore L = {\lim}_{h \to 0} \frac{\cos x - \left(\cos x \cos h - \sin x \sin h\right)}{h \cos \left(x + h\right) \cos x}$
$\therefore L = {\lim}_{h \to 0} \frac{\cos x - \cos x \cos h + \sin x \sin h}{h \cos \left(x + h\right) \cos x}$
$\therefore L = {\lim}_{h \to 0} \frac{\cos x \left(1 - \cos h\right) + \sin x \sinh}{h \cos \left(x + h\right) \cos x}$
$\therefore L = {\lim}_{h \to 0} \frac{\cos x \left(1 - \cos h\right)}{h \cos \left(x + h\right) \cos x} + \frac{\sin x \sinh}{h \cos \left(x + h\right) \cos x}$
 :. L = lim_(h->0) (1 - cos h)/(hcos(x+h)) + (sin x/cos x)lim_(h->0) (sin h)/(hcos(x+h)

As $h \rightarrow 0 \implies \sin h \rightarrow 0 , \cos h \rightarrow 1$
Also, ${\lim}_{h \to 0} \frac{1 - \cos h}{h} = 0$ and ${\lim}_{h \to 0} \left(\sin \frac{h}{h}\right) = 1$

:. L = lim_(h->0) (1 - cos h)/(hcos(x+h)) + (sin x/cos x)lim_(h->0) (sin h)/(hcos(x+h)
$\therefore L = 0 + \left(\sin \frac{x}{\cos} x\right) \frac{1}{\cos} x$
$\therefore L = \tan x \sec x$ QED