How do you show that #lim_(r->oo) e^(ircostheta - rsintheta) = 0#?

1 Answer
Steve M
Sep 4, 2017

# L = lim_(r->oo) e^(ircostheta - rsintheta) #
# \ \ \= lim_(r->oo) e^(ircostheta)e^(- rsintheta) #

Using Euler's formula:

# e^(ix) = cosx + isinx #

We can write:

# L = lim_(r->oo) e^(- rsintheta) { cos(rcostheta) + isin(rcostheta) } #
# \ \ \= lim_(r->oo) e^(- rsintheta) cos(rcostheta) + ilim_(r->oo) e^(- rsintheta) sin(rcostheta) #

Intuitively we can see that the trig portions oscillate between #+-1# but the exponential causes a dampening effect which approaches #0# as #r rarr oo#

Thus:

# L= 0+0i #
# \ \ \= 0 # QED

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