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

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

## marked as duplicate by Jeel Shah, saz, Surb, N. F. Taussig, ChappersMay 7 '15 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. – Andrés E. 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 – Nasibabuba Dec 6 '12 at 20:35
• For positive x's I am done,but negative,I need help, have been trying it for hours... – Nasibabuba 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.

Bernoulli inequality $$1+ny\leq\left(1+y\right)^n$$ using y=x/n we get $$1+x\leq\left(1+\frac{x}{n}\right)^n$$

• Showing Bernoulli's inequality for $-1 < x < 0$ does require some effort though... – Zarrax Dec 6 '12 at 21:06
• @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
• @frhack That's exactly what is written there with $\;\cfrac xn\;$ instead of $\;x\;$ alone. Do you Really have problems to see this, or you just wanted to downvote something that is correct? – DonAntonio Feb 11 '18 at 9:11
• @frhack Don't worry, it happens to us all some time. It is advisable, though, to wait a little more for downvoting even if there's an outstanding mistake in the answer. Sometimes typos go under the radar or someone gets confused. Better, to comment and wait for the answerer to address the doubt. – DonAntonio Feb 11 '18 at 13:35

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.

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.

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 $$e^x > 1+x$$

• 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 $1$? – Andrés E. Caicedo Dec 6 '12 at 21:10
• 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
• What's the problem? $e^x > 0$ for all $x \in \mathbb{R}^n$. – user29999 Oct 8 '13 at 1:24

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.

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 $x > 0$.

• Why the downvote? – JavaMan Dec 7 '12 at 3:18
• I didn't actually downvote (wasn't involved in 2012) but I noticed you proved it for $x>0$ and OP asked for all real $x$ – J. W. Tanner Mar 14 at 0:23

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 $n \to \infty$.

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.

• 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 – Nasibabuba 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