the limit of the sequence is 1 as this is because since b > 1, we have for some

thus: hence therefore i.e.

back to your question:first let such that then thus by definition:

now fix and take of both sides over all rational numbers to get now take of both sides of this

inequality over all rational numbers to get

next let and choose such that and let such that

then: so: taking of both sides over all rational numbers gives

us: now let then using we'll get: which with complete the proof. Q.E.D.

Note:throughout the proof i've assumed that we know all basic properties forrationalpowers of a real number.