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<x<x_0+\delta

e^{x_0-\delta}<e^x<e^{x_0+\delta}

e^{x_0}e^{-\delta}<e^x<e^{x_0}e^{\delta}

e^{x_0}e^{-\delta}-e^{x_0}<e^x-e^{x_0}<e^{x_0}e^{\delta}-e^{x_0}

e^{x_0}(e^{-\delta}-1)<e^x-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}<e^{x_0}(e^{\delta}-1)

|e^x-e^{x_0}|<e^{x_0}(e^{\delta}-1)

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!