Prove that $1+x<e^x$ for all real numbers $x\neq0$.

Question

Prove that $1+x<e^x$ for all real numbers $x\neq0$.

Here is my solution:

We'll prove this as two cases; first when $x<0$ and then when $x>0$.

Let $\alpha>0$. Then, $$e^\alpha-\alpha e^\alpha<1\Rightarrow e^\alpha(1-\alpha)<1\Rightarrow1-\alpha<\frac{1}{e^\alpha}\Rightarrow1-\alpha<e^{-\alpha}.$$ Setting $x=-\alpha$ gives the desired result. Next, consider the fact that $0<\ln(e^\alpha-\alpha)$, as natural log is a strictly positive function. Then, $$0<\ln(e^\alpha-\alpha)\Rightarrow\ln(1)<\ln(e^\alpha-\alpha)\Rightarrow1<e^\alpha-\alpha\Rightarrow1+\alpha<e^\alpha.$$ Setting $x=\alpha$ gives the desired result.

I believe this is a valid proof, but this question is asked in the context of a calculus course, so I'm wondering if there is a different/better way to solve this.

Answer 1

First, $e^x$ is increasing, which is easy to see by computing the derivative.

For $x\gt0$: $\ln(1+x)\lt x\iff\int_1^{1+x}\frac1tdt\lt x\cdot \frac11$ just by considering an upper Riemann sum with one subinterval.

For $-1\lt x\lt0$: $-\ln(1+x)= \int_{1+x}^1\frac1tdt\gt -x\cdot\frac11$, using a lower sum, and the same result follows.

For $x\le -1$, it's obvious, since $e^x\gt0$.

Answer 2

As @Shinalord stated clearly, $\forall x>0$:

$$e^x = 1+x+ \underbrace{\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)}_{>0} > 1 + x$$

Here's a graph:

Answer 3

By Taylor's formula, $$e^x=1+x+\frac{e^\xi}{2}\cdot x^2,$$where $0 \gtrless \xi \gtrless x.$ Hence $$e^x-1-x=\frac{e^\xi}{2}\cdot x^2>0,\forall x \neq 0$$which implies $$e^x>1+x, \forall x \neq 0.$$

Answer 4

Let $f(x)=e^x-x-1$. Then $f'(x)=e^x-1$ and $f''(x)=e^x$. Clearly $f''(x)\gt0$ for all $x$ and $f'(0)=0$, so $f$ has an absolute minimum at $x=0$. I.e., if $x\not=0$, then $f(x)\gt f(0)=0$. Thus $e^x\gt x+1$ for all $x\not=0$.

Answer 5

Another way.

Show that $y=x+1$ is a tangent line to the graph of $f(x)=e^x$ in the point $(0,1)$

and since $f$ is a convex function, we are done!