How can I show $\lim\limits_{x\to a}e^x=e^a$ just using limit ,without "continuous" Effort:
We know that $e^x=\lim\limits_{n\to \infty}\left(1+\dfrac{x}{n}\right)^n$
And if we can say $\lim\limits_{y\to b}\left[\lim\limits_{x\to a}f\right]=\lim\limits_{x\to a}\left[\lim\limits_{y\to b}f\right]$  for this problem;
$\lim\limits_{x\to a}e^x=\lim\limits_{x\to a} \left[\lim\limits_{n\to \infty}\left(1+\dfrac{x}{n}\right)^n\right]=\lim\limits_{n\to \infty}\left[\lim\limits_{x\to a}\left(1+\dfrac{x}{n}\right)^n\right]=\lim\limits_{n\to \infty}\left[\left(1+\dfrac{a}{n}\right)^n\right]=e^a$
and I proof $\boxed{\boxed{\lim\limits_{x\to a}e^x=e^a}}$ .
İs this proof true? And please help ,how I can prove this, with "accurately".
 A: Hint: Use the definition $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ to prove that $e^{x + y} = e^{x}e^{y}$. Then prove that $\lim_{x \to 0}e^{x} = 1$.
In your approach you mention 

And if we can say $\lim\limits_{y\to b}\left[\lim\limits_{x\to a}f\right]=\lim\limits_{x\to a}\left[\lim\limits_{y\to b}f\right]$  for this problem;

How are you sure that we can say this above equality is true for this problem. This kind of equality is not true in general. You need to do further analysis to say that this is true for the current problem.

Update: This is in response to queries from OP in comments. First of all the proof of $e^{x + y} = e^{x}e^{y}$ is non-trivial if we use definition $(1)$. I will skip this for the moment and focus on the easy part of the proof of $\lim_{x \to 0}e^{x} = 1$. First let $x \to 0^{+}$ then we can assume $0 < x < 1$. We have
\begin{align}
e^{x} &= \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\left\{1 + x + \frac{x^{2}}{2!}\left(1 - \frac{1}{n}\right) + \frac{x^{3}}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) + \cdots\right\}\notag\\
& = 1 + x + \lim_{n \to \infty}F(x, n)\text{ (say)}\tag{2}
\end{align}
Now we can see that if $0 < x < 1$ then
\begin{align}
F(x, n) &< \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots + \frac{x^{n}}{n!}\notag\\
&< \frac{x^{2}}{2} + \frac{x^{3}}{4} + \frac{x^{4}}{8} + \cdots\notag\\
&= \frac{x^{2}}{2 - x}\notag
\end{align}
Thus we have $$0 < F(x, n) < \frac{x^{2}}{2 - x}$$ and clearly from definition of $F(x, n)$ we see that $F(x, n)$ is increasing as $n$ increases and by above inequality it is bounded above. Hence $$\lim_{n\to\infty}F(x, n) = F(x)$$ exists for $0 < x < 1$ and clearly $$0 \leq F(x) \leq \frac{x^{2}}{2 - x}$$ and hence by Squeeze Theorem $$\lim_{x \to 0^{+}}F(x) = 0$$ From $(2)$ we have $$e^{x} = 1 + x + F(x)$$ and therefore $\lim_{x \to 0^{+}}e^{x} = 1$.
The proof for $x \to 0^{-}$ requires us to put $x = -y$ and use the fact that $e^{x} = 1/e^{-y}$. This is an easy consequence of $e^{x + y} = e^{x}e^{y}$ which we can prove using the following lemma:

Lemma: If $a_{n}$ is a sequence of real or complex numbers with $n(a_{n} - 1) \to 0$ as $n \to \infty$ then $a_{n}^{n} \to 1$ as $n \to \infty$.

We take the sequence $$a_{n} = \dfrac{\left(1 + \dfrac{x}{n}\right)\left(1 + \dfrac{y}{n}\right)}{\left(1 + \dfrac{x + y}{n}\right)}$$ and see that $n(a_{n} - 1) \to 0$ and hence $a_{n}^{n} \to 1$ and therefore $e^{x}e^{y} = e^{x + y}$.
Proof of the above lemma is available here.
