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} \ $.

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    $\begingroup$ 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. $\endgroup$ – Lee Mosher Jul 21 at 17:11
  • $\begingroup$ 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 $(1+\frac{x}{k})^{k\alpha}$. $\endgroup$ – kolobokish Jul 21 at 17:36
  • $\begingroup$ @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. $\endgroup$ – Gustavo Jul 21 at 17:43
  • $\begingroup$ @Gustavo Please put clarifications of the question into the body of the question, rather than the comments. $\endgroup$ – 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 $1+x<e^x$, 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 $e^x$!

  • $\begingroup$ 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? $\endgroup$ – Gustavo Jul 22 at 20:10
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    $\begingroup$ @Gustavo I just had to do a double take there for a second too, but the justification is correct because we are subtracting these values instead of adding them. I'll write the inequality in terms of the negatives in order to make it more clear. $\endgroup$ – Isaac Browne Jul 22 at 22:08
  • $\begingroup$ Perfect! Thanks $\endgroup$ – Gustavo Jul 22 at 22:52

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