What is the limit as x approaches infinity of $e^{2 x} cos x$ $e^(2x)cosx$ ?

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Answer 1 · 2018-06-14T12:40:11

The limit does not exist.

Consider in fact the two sequences:

$a_{n} = 2 n π$ $a_n = 2npi$

$b_{n} = (2 n + 1) π$ $b_n = (2n+1)pi$

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:

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

What is the limit as x approaches infinity of e^(2x)cosxe2xcosxe^(2x)cosx?