How can you prove e-infinity equals 0?

Answer 1

I think you mean e to the (- infinity) power. The proof would be a limit proof.

The limit (as n-->infinity) of [( en) ] = 0

You should have some other limits in class that you have proven.

Show that your limit is less than one of those given for all values of n then you have your proof.

For instance, if you already know that lim (as n-->infinity) of [(1/n) ] = 0

then for n = 1, 1/e1 < 1/1 true

for n = 2, 1/e2 < 1/2 true

Then assume that it is true for n = k

so for n = K, 1/eK < 1/K assumed true

therefore: eK > K

multiply both by e

e(k+1) > ke

but we know ke > k+1 because e>1

SO: e(k+1) > ke > k+1 now take the reciprocal (reverses the inequalities)

1/e(k+1) < 1/ke < 1/ (k+1)

by transitive prop of inequalities eliminate the middle term

so that 1/e(k+1) < 1/ (k+1) this proves the case for n=K+1 and therefore will be true for all values of n since k was never a specified value.

And if: 1/e(k+1) < 1/ (k+1) by one of the properties of limits, since the lim of 1/n is zero, then the lim of 1/en is also zero when n --> infinity.