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

1 Answer
Steve M
Nov 16, 2016

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

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