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Answer 1 · 2017-10-03T17:33:11

#lim_(n->oo) (1+sin(2/n))^n = e^2#

Write the sequence as:

#(1+sin(2/n))^n = e^(nln(1+sin(2/n))#

Now note that:

#nln(1+sin(2/n)) = 2ln(1+sin(2/n))/sin(2/n) xx sin(2/n)/(2/n)#

If the limits exist, as for #n->oo# we have that #x_n = 2/n->0# and #x'_n = sin(2/n) -> 0# then:

#lim_(n->oo) ln(1+sin(2/n))/sin(2/n) = lim_(x->0) ln(1+x)/x =1#

#lim_(n->oo) sin(2/n)/(2/n) = lim_(x->0)sinx/x =1#

Thus:

#lim_(n->oo) 2ln(1+sin(2/n))/sin(2/n) xx sin(2/n)/(2/n) = 2#

And as #e^x# is a continuous function for all #x in RR#, then:

#lim_(n->oo) e^(nln(1+sin(2/n))) = e^((lim_(n->oo) nln(1+sin(2/n)))) = e^2#