Question

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

Answer 1

The limit does not exist.

Explanation:

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`