How do you find the limit of #sec((pix)/6)# as #x->7#?

Question

3271 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-30T21:44:23

#(-2root2 3)/3 #

First we consider how #1/cosx = secx#

Hence we can say we are finding;

#lim_(x to 7 ) 1/cos((xpi)/6)#

But we know #lim_(x to 7) cos((pix)/6)# = #-root2 3 /2#
Via using a calculator and evaluating when #x=7#

or by cosnidering how #cos(pi/6) = root2 3 /2#
and how

#cos((7pi)/6) = cos(pi + pi/6) = cos(pi)cos(pi/6) - sin(pi)sin(pi/6) #

#-root2 3 /2#
Hence# lim_(x to 7) 1/cos((pix)/6) # = #-2/(root2 3)#

Rationalising the denominator to yeild;

# (-2root2 3 )/3 #