# Prove that $e^x\ge x+1$ for all real $x$ [duplicate]

Without using differentiation, logarithmic function, rigorously, prove that $$e^x\ge x+1$$ for all real values of $x$.

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## marked as duplicate by Jeel Shah, saz, Surb, N. F. Taussig, ChappersMay 7 at 0:06

Please avoid using display math (double \$\$) in titles. –  Jonathan Christensen Dec 6 '12 at 20:28
Are you alowed to use MVT? –  user127.0.0.1 Dec 6 '12 at 20:30
How do you define $e$ and/or $e^x$? What kind of answer is expected depends on the definition you are using. –  Andres Caicedo Dec 6 '12 at 20:31
Yes,I can use MVT, I want to prove it using only lim(1+x/n)n as n goest to inf –  p.s Dec 6 '12 at 20:35
For positive x's I am done,but negative,I need help, have been trying it for hours... –  p.s Dec 6 '12 at 20:37

Bernoulli's Inequality: for any $\,n\in\Bbb N\,$

$$1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x$$

The inequality above is true for $\,x\geq -1\,$ , and since the wanted inequality is trivial for $\,x<-1\,$ we're done.

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Showing Bernoulli's inequality for $-1 < x < 0$ does require some effort though... –  Zarrax Dec 6 '12 at 21:06
@DonAntonio: Nice. Might be clearer if said inequality is trivial for $x\lt -1$. –  André Nicolas Dec 6 '12 at 21:11
@AndréNicolas, that was the intention yet some of my fingers believe they have free will...Thanks. –  DonAntonio Dec 6 '12 at 21:53
@Zarrax, a very minimal effort: the inequality follows by induction in the general case. –  DonAntonio Dec 6 '12 at 21:53
It is quite easy to demonstrate that Bernoull’s inequality is strict for $x\neq 0$ and $n>1$, but the limit introduces it again. Can we extend this to show that the (question) inequality is strict iff $x\neq 0$? –  Steve Powell Sep 25 '13 at 14:26

Write out the expansion of $(1 + {x \over n})^n$ as $$(1 + {x \over n})^n = 1 + n {x \over n} + {n(n-1) \over 2}{x^2 \over n^2} + ...$$ If $x \geq 0$, then all terms are nonnegative, so the sum is at least the sum of the first two terms, namely $1 + x$. If $x \leq - 1$, then for even $n$, the expression $(1 + {x \over n})^n$ is nonnegative, so the limit must be at least zero, which is greater than or equal to $1 + x$.

So it remains to look at the case where $-1 < x < 0$. In this case write $x = -y$ and you have $$(1 - {y \over n})^n = 1 - n {y \over n} + {n(n-1) \over 2}{y^2 \over n^2} - ...$$ This is an (finite) alternating series, and the absolute value of the ratio of two consecutive terms of this series is of the form ${n - k \over k+ 1} {y \over n}$, which is less than $1$ since $0 < y < 1$. So the terms decrease in absolute value. Hence the overall sum is at least what you get if you truncate after a negative term. So truncating after two terms you get $$(1 - {y \over n})^n \geq 1 - y$$ Equivalently, $$(1 + {x \over n})^n \geq 1 + x$$ Taking limits as $n$ goes to infinity gives what you're looking for.

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Hint:

Prove for rational $x=\frac{a}{b}$ ($b > 0$) that $e^{\frac{a}{b}} \ge \frac{a}{b}+1$. You can do this by showing that $$e^a \ge \left(\frac{a}{b}+1\right)^b.$$ Finally, argue by continuity.

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I don't know if it's cheating, but you didn't say that integration is forbidden. Since $\exp$ is increasing we know that $e^{-x} \leq 1$ for all $x \geq 0$. Integrating, we get $$\forall x\geq 0,\qquad 0 = \int_0^x 0\,dt \leq \int_0^x (1-e^{-t})\,dt = x + e^{-x} - 1.$$

The inequality $\forall x \geq 0, \;e^x \geq 1 + x$ follows in the same lines as the former.

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Assuming $x > 0$ and regardless of your definition of $e$, you can show that $e^x = \displaystyle\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n$. Then:

\begin{align} \left( 1 + \frac{x}{n} \right)^n &= \sum_{k=0}^n {n \choose k}\frac{x^k}{n^k} \\ &= 1 + x + \sum_{k=2}^n {n \choose k} \frac{x^k}{n^k} \end{align}

Since each term ${n \choose k} (x/n)^k$ is positive, it follows that

$$\left( 1 + \frac{x}{n} \right)^n - x - 1 > 0.$$ Hence,

$$\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n - x - 1 \geq 0.$$ from which the desired conclusion follows for $x > 0$.

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Why the downvote? –  JavaMan Dec 7 '12 at 3:18

Suppose $x \le 0$. From $t^n - 1 = (t-1)(1 + t + \ldots + t^{n-1})$ (with $t = 1 + x/n$) we get $$(1+x/n)^n - 1 = \frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j$$ If $n$ is large enough that $1+x/n \ge 0$ we have $(1+x/n)^j \le 1$, so $$\sum_{j=0}^{n-1} (1+x/n)^j \le \sum_{j=0}^{n-1} 1 = n$$ and since $x/n \le 0$ $$\frac{x}{n} \sum_{j=0}^{n-1} (1+x/n)^j \ge \frac{x}{n} n = x$$ Thus $$(1+x/n)^n \ge 1 + x$$ Now take the limit as $n \to \infty$.

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Hint: Break it into two cases, $x \leq 0$ and $x > 0$. One of them is trivial. For the other case, consider the Taylor Series expansion.

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Taylor expansion without knowledge about differentiation? –  user127.0.0.1 Dec 6 '12 at 20:29
Knowing a Taylor Series expansion is not the same as using differentiation. Maybe my math education was weird, but I learned Taylor Series in 10th grade, long before I learned differentiation. –  Jonathan Christensen Dec 6 '12 at 20:33
I want to prove it rigourously,Taylor expansion is not allowed –  p.s Dec 6 '12 at 20:36
I think the Taylor series expansion is plenty rigorous, but fair enough. –  Jonathan Christensen Dec 6 '12 at 20:40

If $x\ge 0$ as $\exp$ is increase we have $e^{x} \ge 1$. Then by mean value theorem there exist $c \in (0,x)$ such that $$e^x - 1 = e^c x.$$ Thus $$e^x \ge x +1.$$ If $x \le -1$, we have $$e^x \ge 0 > 1+x$$ Now if $-1 < x < 0$ we have $$e^c < 1 \Rightarrow -e^c > -1 \Rightarrow -e^c x < -x$$ Thus $$1 - e^x = e^c (-x) < -x$$ and $$e^x > 1+x$$

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What if $x \in (-1, 0)$? –  Andrew Uzzell Dec 6 '12 at 21:05
I included this case now, Ok? –  user29999 Dec 6 '12 at 21:08
How is the negative number $-e^c$ larger than $1$? –  Andres Caicedo Dec 6 '12 at 21:10
Thank you, you are right. I corrected this. –  user29999 Dec 6 '12 at 21:15
Your 3rd equation, $e^x \ge 0 > 1+x$ is incorrect. $1 + x$ is zero when $x = -1$ so the second inequality needs to be $\ge$ –  Alexis Wilke Sep 25 '13 at 21:52