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

We want to prove that

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

To prove that ( \lim_{h \to 0} \frac{\sec(x+h) - \sec(x)}{h} = \sec(x) \tan(x) ), we can use the definition of the secant function and the limit properties.

First, recall the definition of secant function: ( \sec(x) = \frac{1}{\cos(x)} ).

Now, rewrite the expression as follows: [ \lim_{h \to 0} \frac{\frac{1}{\cos(x+h)} - \frac{1}{\cos(x)}}{h} ]

Next, find a common denominator: [ \lim_{h \to 0} \frac{\cos(x) - \cos(x+h)}{h \cos(x) \cos(x+h)} ]

Apply the trigonometric identity ( \cos(a) - \cos(b) = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) ): [ \lim_{h \to 0} \frac{-2 \sin\left(\frac{x + (x+h)}{2}\right) \sin\left(\frac{x - (x+h)}{2}\right)}{h \cos(x) \cos(x+h)} ]

Simplify: [ \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(-\frac{h}{2}\right)}{h \cos(x) \cos(x+h)} ]

[ \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(-\frac{h}{2}\right)}{h \cos(x) \left(\cos(x) \cos\left(\frac{h}{2}\right) - \sin(x) \sin\left(\frac{h}{2}\right)\right)} ]

[ \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right) \sin\left(-\frac{h}{2}\right)}{h \left(\cos^2(x) - \sin^2(x)\right)} ]

[ \lim_{h \to 0} \frac{2 \sin\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h \left(\cos^2(x) - \sin^2(x)\right)} ]

Use the trigonometric limit ( \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1 ): [ \frac{2 \cdot 1 \cdot 0}{\left(\cos^2(x) - \sin^2(x)\right)} ]

[ \frac{0}{\cos^2(x) - \sin^2(x)} = 0 ]

Therefore, ( \lim_{h \to 0} \frac{\sec(x+h) - \sec(x)}{h} = \sec(x) \tan(x) ).