How do you find the limit of #sec(2x)# as #x->0#?
1 Answer
Dec 25, 2016
Explanation:
#=lim_(x-> 0) 1/cos(2x)#
#=1/cos(2 * 0)#
#= 1/cos(0)#
#= 1/1#
#=1#
Hopefully this helps!
