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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.

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    @Peter Tamaroff If the sequence x_n converges to x_0, then f(x_n) will tend to f(x_0) as n tends to infinity.2012-12-08

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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| 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.

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    I'm afraid I still don't understand. How exactly have you proven sequential continuity?2012-12-09