How to test continuty of this function ? f(x) = #(1+x)^(1+x)# x is not equal zero

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2 Answers
Andrea S.
Sep 29, 2017

#f(x)# is continuous for every #x in (-1,+oo)#.

Explanation:

Write the function as:

#f(x) = e^((1+x)ln(1+x))#

As the exponential function #e^t# is continuous for any #t in RR#, and #(1+x)ln(1+x)# is continuous for #x > -1# then #f(x)# is continuous for every #x in (-1,+oo)# and #x!=0#

For #x=0#:

#lim_(x->0) f(x) = lim_(x->0) e^((1+x)ln(1+x)) = e^((lim_(x->0)(1+x)ln(1+x))) = e^0 =1 = f(0)#

So the function is continuous also for #x = 0#

graph{(1+x)^(1+x) [-10, 10, -5, 5]}

yumean1119
Sep 29, 2017

This function is not continuous at #x=0#

Explanation:

Consider a function #g(h)=(1+1/h)^h#.

This function approrches #e(=2.718…)# when #h#-> #oo# (by definition of #e#). This is also true when #h# goes to negative infinity.

So, the limit of #f(x)# when #x# approaches 0 is
#lim_(x->+0) f(x)# = #lim_(h->oo) g(h)# =#e#
#lim_(x->-0) f(x)# = #lim_(h->-oo) g(h)# =#e#

They are different from #f(0)# and therefore #f(x)# has a jump
discontinuity at #x=0#.

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