Ytukyg

I need to prove that $e^x ge 1+x$ for every real $x$ by using Bernoulli's inequality.

Hint:

$$(1+m)^n geq 1+mn$$

From here, let

$$m = frac{x}{n}$$

and recall that $big(1+frac{x}{n}big)^n to e^x$ as $n to infty$.

You can also use the solution proposed above, considering

$$e geq bigg(1+frac{1}{n}bigg)^n implies e^x geq bigg(1+frac{1}{n}bigg)^{nx} geq 1+x$$

and including that

$$bigg(1+frac{1}{n}bigg)^n = {n choose 0}+{n choose 1}bigg(frac{1}{n}bigg)+{n choose 2}bigg(frac{1}{n}bigg)^2+…$$

$$= 1+frac{n!}{1!(n-1)!n}+frac{n!}{2!(n-2)!n^2}+…$$

and since all the terms are contain the same degree of $n$ in the numerator and denominator, $n to infty$ yields

$$= 1+1+frac{1}{2!}+frac{1}{3!}+frac{1}{4!}+… to e$$

Since $left(1+frac1nright)^{n}le e$, we have that

$$e^xge left(1+frac1nright)^{xn}stackrel{B.I.}ge 1+nxcdot frac1n=1+x$$

Refer also to the related

• How to prove that $lim_{ntoinfty}big(1+1/nbig)^n$ is equal to e

Bernoulli's inequality:

1) $(1+y)^n ge 1+ny$, where $n in mathbb{N}$, and $y ge -1$.

Let $x$ be real, then for large enough $n$:

$|x| <n$ , or $|x|/n<1$.

We then have $(1+x/n)^n ge 1+x$.

2) $e^x = lim_{n rightarrow infty}(1+x/n)^n$, $x$ real.

The sequence $(1+x/n)^n$ is increasing

for $n >|x|$. $e^x$ is an upper bound

(Proving that sequence $x_n=(1+x/ n)^n$ is increasing and bounded for $n>|x|$ and therefore the $lim_{n to infty}x_n$ exists).

Finally:

$e^x > (1+x/n)^n >1+x$.

Let the exponential function be defined as
$$exp(x)=e^x=lim_{ntoinfty}Big(1+frac{x}{n}Big)^n$$
Step 1:
First we prove $exp(x)geq 0$ for all $xinmathbb{R}$.

For an arbitrary $xinmathbb{R}$ there is a natural number $n$ such that $|x|<n$, thus $-1<frac{x}{n}<1$. This implies $0<1+frac{x}{n}<2$, thus
$$Big(1+frac{x}{n}Big)^n>0$$
For all $Ngeq n$ we have then
$$Big(1+frac{x}{N}Big)^N>0$$
Thus
$$lim_{Ntoinfty}Big(1+frac{x}{N}Big)^N=e^xgeq 0$$

Step 2: Next we show $e^xgeq 1+x$ for all $xinmathbb{R}$.

• The inequality is obviously true for $xleq -1$, since the right hand side is smaller or equal to zero, but the left hand side is greater or equal to zero.


• Now consider the case $x>-1$. Then for all $ninmathbb{N}$ we have $frac{x}{n}>-1$ and finally we use the Bernoulli's inequality to obtain
  $$Big(1+frac{x}{n}Big)^ngeq 1+ncdot frac{x}{n}=1+x$$
  Sending $ntoinfty$ you get the result.