[Math] Prove that the exponential function is sequentially continuous

Question

I am supposed to use the definition of the exponential function to prove that if x is a real number and the modulus of x is less than 1, the modulus of exp(x)-1 is less than or equal to (e-1)*modulus of x, and hence prove that the exponential function is sequentially continuous. I've managed to prove the former using the exponential power series but I don't understand how to 'hence' prove the latter.

Answer 1

You can prove this in several ways:

$(1)$ The exponential function is continuous. A function is continuous at ($x=x_0$) $\iff$ it is sequentially continuous (at $x=x_0$), thus the result follows.

$(2)$ The exponential function is differentiable over $\Bbb R$. Given a convergent sequence $\{a_n\}$, the sequence will be bounded over some interval $[a,b]$. Then we will have

$$|\exp(x_n)-\exp(x_0)|=\exp(\chi)|x_m-x_0|<M|x_m-x_0|$$ by the mean value theorem where $M$ is the maximum of $\exp$ (which is $ =\exp'$) over $[a,b]$.

Note that the exponential function has the superb property that being continuous at $x=0$ immediately means it is continuous at every other real. This is because of the functional equation $$\exp(x+y)=\exp(x)\exp(y)$$

Indeed, making $$\exp(x+h)-\exp(x)$$ small accounts to making $$\exp(x)(\exp(h)-1)$$

small. This can happen if $\exp(h)\to 1 $ when $h\to 0$ which is indeed true and means $\exp(h)=\exp(0)=1$ whence $\exp(h)$ is continuous at $x=0$. Thus, all you need is to show, as you're being asked, is that the exponential function is continuous at $x=0$, which means that if $|h|$ is small, so is $|e^h-1|$. My way would be to argue that since

$$1=\log e=\lim_{h\to 0}\frac{e^h-1}{h}$$ then

$$\lim_{h\to 0}]({e^h-1})=\lim_{h\to 0}h\frac{e^h-1}{h}=0\cdot 1=0$$ and continuity follows.