# Proving $1+x^{\alpha} < e^{\alpha x}$ without calculus

Let $$\ \alpha , x \in \mathbb{R} \$$ be such that $$\ \alpha \geq 1 \$$ and $$\ x>0 \$$. Show that $$1+x^{\alpha} < e^{\alpha x}$$ without using the tools of Calculus (beginning on differentiation) or more advanced mathematics.

I have a clue how to solve it using derivatives and the Taylor series, but I can only use basic properties of limits of functions, I can use nothing that comes from derivatives, integrals, infinite series and other things like that. I tried to approach it via the definition of the number $$\ e = \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right)^k \$$ and the Bernoulli's inequality, but I failed.

Any help is appreciated.

EDIT: Like I said in the comments, the definition is $$e^s = \lim_{k \to \infty} \left( 1 + \frac{s}{k} \right)^k \ \ \ ,$$ for all $$\ s \in \mathbb{R} \$$.

• This seems like a hopeless task since the very definition of $e^{\alpha x}$ uses calculus. Unless you have some other definition in mind? In that case, you should tell us what definition you are working from. – Lee Mosher Jul 21 at 17:11
• Bernoulli's inequality will help. Namely $(1+y)^n \geq 1+ny$. Make some changes and this will do. (or a strict version of it with $n \geq 2$). Some hint if you need it consider $(1+\frac{x}{k})^{k\alpha}$. – kolobokish Jul 21 at 17:36
• @LeeMosher the definition is $\ e^{\alpha x} = \lim_{k \to \infty} \left( 1 + \frac{\alpha x}{k} \right)^k \$ and I can use only basic properties of limits. – Gustavo Jul 21 at 17:43
• @Gustavo Please put clarifications of the question into the body of the question, rather than the comments. – saulspatz Jul 21 at 17:49

First, we can start with the inequality $$1+x^\alpha \leq (1+x)^\alpha$$ for $$x>0, \alpha\geq 1$$. For $$\alpha=1$$, this is obvious as equality holds, and otherwise, we can write \begin{align*} 1 &= (1+x) -x & \text{obvious}\\ 1&\leq(1+x)^{\alpha-1}[(1+x)-x] &\text{since } (1+x)^{\alpha-1}>1\\ 1&\leq(1+x)^\alpha - (1+x)^{\alpha-1}x < (1+x)^\alpha-x^\alpha &\text{since } -(1+x)^{\alpha-1}<-x^{\alpha-1}\\ 1+x^\alpha &<(1+x)^\alpha &\text{rearranging} \end{align*} Then, we just have to prove $$1+x, which can be done by using your limit definition, that $$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ And since we have $$\left(1+\frac{x}{n}\right)^n<\left(1+2\frac{x}{2n}+\frac{x^2}{4n^2}\right)^n=\left(1+\frac{x}{2n}\right)^{2n}$$ We can write $$1+x<\left(1+\frac{x}{2}\right)^2<\cdots \to e^x$$ And so we're done, as we know $$1+x$$ is the beginning of an increasing sequence which tends to $$e^x$$!
• I did not understand step $(1+x)^{\alpha} - (1+x)^{\alpha-1}x < (1+x)^{\alpha} - x^{\alpha}$. The justification is written $(1+x)^{\alpha - 1} > x^{\alpha - 1}$, but it seems I need the opposite of this to follow the inequalities. Am I wrong? – Gustavo Jul 22 at 20:10