What is the limit as x approaches infinity of #e^(2x)cosx#?

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Answer 1 · 2018-06-14T12:40:11

The limit does not exist.

Consider in fact the two sequences:

#a_n = 2npi#

#b_n = (2n+1)pi#

In both cases:

#lim_(n->oo) a_n = lim_(n->oo) b_n = +oo#

If the limit:

#lim_(x->oo) e^(2x)cosx #

existed then it would coincide with #lim_(n->oo) e^(2a_n)cos(a_n) # and #lim_(n->oo) e^(2b_n)cos(b_n)#.

But:

#lim_(n->oo) e^(2a_n)cos(a_n) = lim_(n->oo) e^(4pin)cos(2pin) = lim_(n->oo) e^(4pin) = +oo#

while:

#lim_(n->oo) e^(2b_n)cos(b_n) = lim_(n->oo) e^(4pin+2pi)cos((2n+1)pi) = -e^(2pi)lim_(n->oo) e^(4pin) = -oo#