Question

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

Answer 1

See belw

Explanation:

We want to prove that

`lim_(h->0)(sec(x+h) - sec x)/h = sec x*tan x`

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

`d/dx sec(x) = sec x*tan x`

Let `L = lim_(h->0)(sec(x+h) - sec x)/h`, Then:

`L = lim_(h->0)(1/(cos(x+h)) - 1/(cos x))/h`
`:. L = lim_(h->0) (cos x - cos(x+h))/(hcos(x+h)cos x)`
`:. L = lim_(h->0)(cos x - (cos xcos h - sin xsin h))/(hcos(x+h)cos x)`
`:. L = lim_(h->0)(cos x - cos xcos h + sin xsin h)/(hcos(x+h)cos x)`
`:. L = lim_(h->0) (cos x(1 - cos h)+sinxsinh)/(hcos(x+h)cos x)`
`:. L = lim_(h->0) (cos x(1 - cos h))/(hcos(x+h)cos x) + (sin xsinh)/(hcos(x+h)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 rarr 0 => sin h rarr 0, cos h rarr 1`
Also, `lim_(h->0)(1 - cos h)/h = 0` and `lim_(h->0) (sin h / h) = 1`

`:. L = lim_(h->0) (1 - cos h)/(hcos(x+h)) + (sin x/cos x)lim_(h->0) (sin h)/(hcos(x+h)`
`:. L = 0+ (sin x/cos x)1/cosx`
`:. L = tanxsecx` QED