F(x)=(e^1/x +1)/(e^1/x -1) is function continuous at x=0?

Question

Do you mean #F(x)=(e^(1/x) +1)/(e^(1/x) -1)#

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Answer 1 · 2017-12-15T12:07:50

No. Lateral limits are different at #x_0=0#

#lim_(xrarr0^-)f(x)# #=# #lim_(xrarr0^-)(e^(1/x)+1)/(e^(1/x)-1)# #=-1#

#lim_(xrarr0^+)f(x)# #=# #lim_(xrarr0^+)(e^(1/x)+1)/(e^(1/x)-1)# #= 1#

so #lim_(xrarr0^-)f(x)##!=##lim_(xrarr0^+)f(x)# and #f# is as a result not continuous at #x_0=0#

#lim_(xrarr0^-)(e^(1/x)+1)/(e^(1/x)-1)#

#1/x=y#
#x->0^-#
#y->-oo#

#=# #lim_(yrarr-oo)(e^y+1)/(e^y-1)# #= (0+1)/(0-1) = -1#

#lim_(xrarr0^+)(e^(1/x)+1)/(e^(1/x)-1)#

#1/x=u#
#x->0+#
#u->+oo#

#=# #lim_(urarr+oo)(e^u+1)/(e^u-1)# -- #((+oo)/(+oo))#

and with rules De L'Hopital

#=# #lim_(urarr+oo)(e^u)/(e^u)# #=# #lim_(urarr+oo)1# #=1#

(Note: you can check in the graph the behaviour of #e^x# while #x->+-oo# )

graph{e^x [-22.79, 22.82, -11.42, 11.37]}

Answer 2 · 2017-12-15T15:18:34

#f(x) = e^(1/(x+1))/e^(1/(x-1))# is continuous at #0#.

Both #1/(x+1)# and #1/(x-1)# are continuous at #0#, so

both #e^(1/(x+1))# and #e^(1/(x-1))# are continuous at #0#

Because #e^(1/(x-1)) != 0# at #x=0#,

the quotient #e^(1/(x+1))/e^(1/(x-1))# is also continuous at #0#.