How to prove that #f# is continuous?

#f,g:R->R# with
#g(x)=f(f(x)) + e^x# ,#xεR#
supposed #f#, #g# are strictly increasing and for #x_0εR#, #lim_(xrarrx_0)f(x)=l#

1 Answer
Dec 5, 2017

#lim_(xrarrx_0)f(x)=f(x_0)=L#

Explanation:

#lim_(xrarrx_0)f(x)=L# #<=>##lim_(xrarrx_0^+)f(x)#=#lim_(xrarrx_0^-)f(x)=L#

  • If #x<##x_0# #=># #x->x_0^-# #=>##f(x)##<##f(x_0)# , because #f# strictly increasing

so #=># #lim_(xrarrx_0^-)f(x)# #<=##f(x_0)# #(1)#

  • If #x>##x_0# #=># #=># #x->x_0^+# #=>##f(x)##>##f(x_0)# , because #f# strictly increasing

so #=># #lim_(xrarrx_0^+)f(x)# #>=##f(x_0)# #(2)#

#L<=f(x_0)#
&
#L>=f(x_0)#

#=># #f(x_0)=L#