elementary proof of growth of exponential function
Proposition 1.
If is a non-negative real number and is a non-negative integer, then .
Proof.
When , we have .If, for some natural number , it is the case that then, multiplying both sides ofthe inequality
by , we have
By induction, for everynatural number .∎
Proposition 2.
If is a real number such that and and are non-negative integers, we have.
Proof.
Let . Write where and are non-negative integers and .
By the preceding proposition, . Raising bothsides of this inequality to the power, wehave . Since , we alsohave ; multiplying bothsides by this inequality and collecting terms,
Multiplying the right-hand side by andrearranging,
Since , we also have
Recalling that and , we conclude that
∎
Proposition 3.
If , , and are real numbers such that , and , then
Proof.
Let and be integers such that and . Since , we have . By the preceeding proposition, we have
Since , we have , so
Since , we have
Summarrizing our progress so far,
Dividing both sides by and simplifying,
∎
Proposition 4.
If and are real numbers and , then
Proof.
Substituting for
Dividing by and rearranging,
Since and , we also have by the squeeze rule.
∎