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".

  • $\begingroup$ Are you asking about whether we can say $\lim_{y\rightarrow b}\lim_{x\rightarrow a}f = \lim_{x\rightarrow a}\lim_{y\rightarrow b} f$? Because that's not a legal manipulation in general. Certainly, your proof works if we can say that, but that might not be so meaningful given that we can't say that. $\endgroup$ – Milo Brandt Jun 4 '16 at 16:39
  • $\begingroup$ yes you right,not everytime, counter example : if $f\equiv f(x,y)$ $(x^2-y^2)/(x^2+y^2)$ and $(x,y)=(0,0)$ $\endgroup$ – pHotone Jun 4 '16 at 16:43
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    $\begingroup$ In this case the conditions for interchanging limits comes down to showing that $e^x$ is continuous. $\endgroup$ – Henrik supports the community Jun 4 '16 at 16:51
  • $\begingroup$ @Henrik All conditions functions ,can interchanging like that? $\endgroup$ – pHotone Jun 4 '16 at 16:59
  • $\begingroup$ Did you mean "continuous" rather than "conditions"? $\endgroup$ – user247327 Jun 4 '16 at 17:26

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.

  • $\begingroup$ Others I did but last one ,Iam not sure how proove $\lim_{x \to 0}e^{x} = 1$ that's very clearly but how?; $e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$ $e^{x+y}= \lim_{n \to \infty}\left(1 + \frac{x+y}{n}\right)^{n}=\left[\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\right].\left[\lim_{n \to \infty}\left(1 + \frac{y}{n}\right)^{n}\right]$ $\lim\limits_{x\to 0}e^{x} =\underbrace{\lim\limits_{x\to 0}\left[\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\right]}_{e^0}=1$ $\endgroup$ – pHotone Jun 4 '16 at 16:58
  • $\begingroup$ " you need to justify the interchange of limits via some further analysis." What's this exactly mean?(sorry for my badly englısh :) ) $\endgroup$ – pHotone Jun 4 '16 at 16:58
  • $\begingroup$ @Kekoçiçeskiye: will provide reply in my updated answer. wait for some time. $\endgroup$ – Paramanand Singh Jun 4 '16 at 16:59
  • $\begingroup$ @Kekoçiçeskiye: Please see updated answer. $\endgroup$ – Paramanand Singh Jun 4 '16 at 17:28
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    $\begingroup$ Paramanand. Just to clarify, Bernoulli's Inequality isn't my favorite tool. But, I know what you mean. $\endgroup$ – Mark Viola Jun 5 '16 at 3:50

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