Question

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

Answer 1

`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