How do you find the limit of #1-2^-(1/x)/(1+2^-(1/x))# as x approaches #0^-#?

Question

1311 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-13T09:26:58

#lim_(x->0^-)[1-2^-(1/x)/(1+2^-(1/x))]=0#

We can write #1-2^-(1/x)/(1+2^-(1/x))#

as #1-1/(2^(1/x)+1)# (multiplying numertaor and denominator by #2^(1/x)#)

When #x->0^-#, #2^(1/x)->0#

Hence as #x->0^-#

#1-2^-(1/x)/(1+2^-(1/x))->1-1/(1+0)=0#

Hence #lim_(x->0^-)[1-2^-(1/x)/(1+2^-(1/x))]=0#

Observe that as #x->0^+#, as #1-2^-(1/x)/(1+2^-(1/x))=1-1/(2^(1/x)+1)#

as #2^(1/x)->oo#, #1-2^-(1/x)/(1+2^-(1/x))->1-0=1#

graph{1-2^-(1/x)/(1+2^-(1/x)) [-2.5, 2.5, -1.25, 1.25]}