# elementary proof of growth of exponential function

###### Proposition 1.

If $x$ is a non-negative real number and $n$ is a non-negative integer, then $(1+x)^{n}\geq 1+nx$.

###### Proof.

When $n=0$, we have $(1+x)^{0}=1\geq 1+0\cdot 0$. If, for some natural number  $n$, it is the case that $(1+x)^{n}\geq 1+nx$ then, multiplying both sides of the inequality  by $(1+x)$, we have

 $(1+x)^{n+1}\geq(1+x)(1+nx)=1+(n+1)x+nx^{2}\geq 1+(n+1)x.$

By induction  , $(1+x)^{n}\geq 1+nx$ for every natural number $n$. ∎

###### Proposition 2.

If $b$ is a real number such that $b>1$ and $n$ and $k$ are non-negative integers, we have $b^{n}>(\frac{b-1}{bk})^{k}n^{k}$.

###### Proof.

Let $x=b-1$. Write $n=mk-r$ where $m$ and $r$ are non-negative integers and $r.

By the preceding proposition   , $(1+x)^{m}>mx$. Raising both sides of this inequality to the $k^{\hbox{th}}$ power, we have $(1+x)^{mk}>(mx)^{k}$. Since $r, we also have $(1+x)^{-r}>(1+x)^{-k}$; multiplying both sides by this inequality and collecting terms,

 $(1+x)^{mk-r}>(\frac{x}{1+x})^{k}m^{k}.$

Multiplying the right-hand side by $k^{k}/k^{k}$ and rearranging,

 $(\frac{x}{1+x})^{k}m^{k}=(\frac{x}{(1+x)k})^{k}(mk)^{k}.$

Since $mk\geq mk-r$, we also have

 $(\frac{x}{(1+x)k})^{k}(mk)^{k}\geq(\frac{x}{(1+x)k})^{k}(mk-r)^{k}.$

Recalling that $mk-r=n$ and $1+x=b$, we conclude that

 $b^{n}>(\frac{b-1}{bk})^{k}n^{k}.$

###### Proposition 3.

If $a$, $b$, and $x$ are real numbers such that $a\geq 0$, $b>1$ and $x>0$, then

 $b^{x}>(\frac{(b-1)^{a}}{b^{a+1}(a+1)^{a}})x^{a}.$
###### Proof.

Let $k$ and $n$ be integers such that $a\leq k\leq and $x\leq n\leq x+1$. Since $x+1>n$, we have $b^{x+1}>b^{n}$. By the preceeding proposition, we have

 $b^{n}>(\frac{b-1}{bk})^{k}n^{k}.$

Since $k, we have $1/k^{k}>1/(a+1)^{k}$, so

 $(\frac{b-1}{bk})^{k}>(\frac{b-1}{b(a+1)})^{k}.$

Since $k\geq a\geq 0$, we have

 $(\frac{b-1}{b(a+1)})^{k}n^{k}\geq(\frac{b-1}{b(a+1)})^{a}n^{a}.$

Summarrizing our progress so far,

 $b^{x+1}>(\frac{b-1}{b(a+1)})^{a}n^{a}.$

Dividing both sides by $b$ and simplifying,

 $b^{x}>(\frac{(b-1)^{a}}{b^{a+1}(a+1)^{a}})x^{a}.$

###### Proposition 4.

If $a$ and $b$ are real numbers and $b>1$, then

 $\lim_{x\to\infty}\frac{x^{a}}{b^{x}}=0.$
###### Proof.

Substituting $a+1$ for $a$

 $b^{x}>(\frac{(b-1)^{a+1}}{b^{a+2}(a+2)^{a+1}})x^{a+1}.$

Dividing by $x$ and rearranging,

 $0<\frac{x^{a}}{b^{x}}<(\frac{b^{a+2}(a+2)^{a+1}}{(b-1)^{a+1}})\frac{1}{x}$

Since $\lim_{x\to\infty}0=0$ and $\lim_{x\to\infty}\frac{1}{x}=0$, we also have $\lim_{x\to\infty}\frac{x^{a}}{b^{x}}=0$ by the squeeze rule.

Title elementary proof of growth of exponential function ElementaryProofOfGrowthOfExponentialFunction 2014-03-10 17:57:26 2014-03-10 17:57:26 rspuzio (6075) rspuzio (6075) 28 rspuzio (6075) Definition msc 26A12 msc 26A06