Which values of x this function is continuous? (x non-negative real number and n->infinity)

#f(x)=limlog(e^n+x^n)/n#

2 Answers
Jan 5, 2018

For #x in (0,e)#, write the function under the limit as:

#log (e^n+x^n)/n = log (e^n(1+(x/e)^n))/n#

and using the properties of logarithms:

#log (e^n+x^n)/n = (log (e^n)+ log(1+(x/e)^n))/n#

#log (e^n+x^n)/n =1+ ( log(1+(x/e)^n))/n#

Now, as #x/e < 1# we have #lim_(n->oo) (x/e)^n = 0#, so #f(x) = 1#

For #x = e#, we have that:

#log (e^n+x^n)/n = log (2e^n)/n = (log2+n)/n#

and again #f(x) = 1#

For #x > e# write the function as:

#log (e^n+x^n)/n = log (x^n((e/x)^n+1))/n#

#log (e^n+x^n)/n = (nlogx+log((e/x)^n+1))/n#

#log (e^n+x^n)/n = logx+log((e/x)^n+1)/n#

We have now #e/x < 1# and # lim_(n->oo) (e/x)^n =0#, so #f(x) = logx#

We can conclude that the function #f(x)# is:

#f(x) = {(1 " for " x <=e),(logx " for " x > e):}#

and is therefore continuous for every #x in RR#

Jan 5, 2018

It is continuous for all non-negative real #x#

Explanation:

It looks like the function is:

#f(x) = lim_(nrarroo)ln(e^n+x^n)/n#

In the following #x# always represents a non-negative real number.

For every #x# #lim_(nrarroo)(e^n+x^n) = oo# so the defining limit above has indeterminate form #oo/oo#.

Apply l"Hospital's Rule (differentiate with respect to #n#) and find

#lim_(nrarroo)(e^n+x^nlnx)/(e^n+x^n)#

Case 1 For #x < e# so #x/e < 1# divide numerator and denominator by #e# to get

#(1+(x/e)^nlnx)/(1+(x/e)^n# whose limit at #oo# is #1#.

Case 2 For #x=e#, we have #lim_(nrarroo) (e^n+e^n)/(e^n+e^n) = lim_(nrarroo)1 = 1#

Case 3 For #x > e# so #e/x < 1# divide numerator and denominator by #e# to get

#((e/x)^n+lnx)/((e/x)^n+1)# whose limit at #oo# is #lnx#.

So,

#f(x) = {(1,"if",x<=e),(lnx,"if",x>e):}# which is continuous at all #x >=0#