## Continuity

Prove that $f(x)=e^x$ is continuous.
This is what i've done so far:

The definition of continuity is $\forall \ \epsilon>0 \ \exists \ \delta>0 \ s.t \ |x-x_0|< \delta \ \Rightarrow \ |f(x)-f(x_0)|< \epsilon$

In this case, $|f(x)-f(x_0)|< \epsilon$ will be $|e^x-e^{x_0}|< \epsilon$

$|x-x_0|< \delta$

$x_0-\delta

$e^{x_0-\delta}

$e^{x_0}e^{-\delta}

$e^{x_0}e^{-\delta}-e^{x_0}

$e^{x_0}(e^{-\delta}-1)

Since $\delta>0$, $e^{-\delta} \geq -e^{\delta}$

Thus:

$e^{x_0}(e^{-\delta}-1)>-e^{x_0}(e^{\delta}+1)>-e^{x_0}(e^{\delta}-1)$

Combining this chain of inequalities with the one before gives:

$-e^{x_0}(e^{\delta}-1)

$|e^x-e^{x_0}|

Therefore $\forall \ \epsilon>0$ let $\epsilon=e^{x_0}(e^{\delta}-1)$

Hence continuity is proved. I think this is correct, I would just like someone to look over it and tell me if this is how they would have written it or if this is even remotely correct.

Here's my other attempt with sequences:

Let $a_n=x_0+\frac{1}{n}$ and let $b_n=x_0-\frac{1}{n}$.

As $n \rightarrow \infty$, $a_n$ approaches $x_0$ from above $(\lim_{n \rightarrow \infty}a_n=\lim_{a_n \rightarrow x_0+}f(a_n))$ and $b_n$ approaches $x_0$ from below $(\lim_{n \rightarrow \infty}b_n=\lim_{b_n \rightarrow x_0-}f(b_n))$.

Therefore:

$f(\lim_{n \rightarrow \infty} a_n)=\lim_{n \rightarrow \infty}e^{x_0+\frac{1}{n}}=e^{x_0}$

$f(\lim_{n \rightarrow \infty} b_n)=\lim_{n \rightarrow \infty}e^{x_0-\frac{1}{n}}=e^{x_0}$

The right and left hand limits are the same so the function is continuous $\forall \ x_0 \in \mathbb{R}$.

My other query with this is that I have only shown that the function is continuous for one set of sequences. Is there a way of showing that the function is continuous for all sequences that converge to $x_0$?

Thanks in advance to anyone who posts!