The limit definition of differentiation is determined by applying the property below
color(red)(f'(x)=lim_(h->0)(f(x+h)-f(x))/h)
f'(x)=lim_(h->0)(sec(x+h)-secx)/h
f'(x)=lim_(h->0)(1/cos(x+h)-1/cosx)/h
f'(x)=lim_(h->0)((cosx-cos(x+h))/(cos(x+h)cosx))/h
f'(x)=lim_(h->0)((2sin((x+x+h)/2)xxsin((x+h-x)/2))/(cos(x+h)cosx))/h
f'(x)=lim_(h->0)((2sin((2x+h)/2)xxsin(h/2))/(cos(x+h)cosx))/h
f'(x)=lim_(h->0)(2sin((2x+h)/2)xxsin(h/2))/(hcos(x+h)cosx)
f'(x)=lim_(h->0)(2sin(x+h/2)xxsin(h/2))/((color(blue)2h)/color(blue)2cos(x+h)cosx)
f'(x)=lim_(h->0)(2sin(x+h/2)xxsin(h/2))/(2(h/2)cos(x+h)cosx)
f'(x)=lim_(h->0)color(brown)(sin(h/2)/(h/2))xx(2sin(x+h/2))/(2cos(x+h)cosx)
Knowing that :
color(brown)(lim_(u->0)sinu/u=1)
So,
color(brown)(lim_(h->0)sin(h/2)/(h/2)=lim_(h/2->0)sin(h/2)/(h/2)=1
Then the limit as h->0 is
f'(x)=lim_(h->0)color(brown)1xx(cancel2sin(x+0/2))/(cancel2cos(x+0)cosx)
f'(x)=sinx/(cosxxxcosx)
Therefore,
f'(x)=sinx/cos^2x
