# Limit involving the exponential of an expression plus a small-o

I'm trying to formalize the following limit:

$$\lim_{x \to \infty} \left[1+\frac{a}{x}+o\left(\frac{1}{x}\right) \right]^x= e^a$$ however I can't see how to rigorously show this identity. Any suggestions?

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Try showing the logarithm of this identity. – Qiaochu Yuan Oct 11 '11 at 22:19
For any $\epsilon>0$, can you show that it is greater than $e^{a-\epsilon}$ and less than $e^{a+\epsilon}$? – robjohn Oct 11 '11 at 22:25
Ok, I've solved it using @QiaochuYuan's hint, re-expanding with Taylor the logarithm etc. Thanks :) – Immanuel Weihnacht Oct 11 '11 at 22:29
@robjohn Yep, now I see... I had just to apply the definition... I have to be ashamed by the mental lazyness that brings me to ask somebody before to ask myself one more time... Thank you – Immanuel Weihnacht Oct 11 '11 at 22:38
the one trickiness about using my hint is that you should use $\liminf$ and $\limsup$. – robjohn Oct 11 '11 at 22:49

Since you've solved it using Qiaochu's logarithm hint, let's try another way. I am guessing that you already know that $$\lim_{x\to\infty}\left[1+\frac{a}{x}\right]^x=e^a\tag{1}$$ and that the new twist is the little-o term. Little-o here means that for any $\epsilon>0$, there is an $M>0$ so that for $x>M$ $$\left|o\left(\frac{1}{x}\right)\right|\le\frac{\epsilon}{x}\tag{2}$$ Condition $(2)$ means that for $x>M$ $$\left[1+\frac{a-\epsilon}{x}\right]^x\le\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le\left[1+\frac{a+\epsilon}{x}\right]^x\tag{3}$$ Inequalities $(3)$ imply that $$e^{a-\epsilon}\le\liminf_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le\limsup_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le e^{a+\epsilon}\tag{4}$$ and since inequalities $(4)$ hold for any $\epsilon>0$, we have that $$\lim_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x=e^a$$ Note that we cannot apply the Squeeze Theorem to $(3)$ because the limits of the upper and lower bounds are not equal. Since $\epsilon>0$, $e^{a-\epsilon}\not=e^{a+\epsilon}$; however, since $\epsilon$ can be chosen arbitrarily small, $(4)$ allows us to squeeze the $\liminf$ and $\limsup$ together to produce a limit.

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I am confused about the need for the inferior and superior limits. After attaining (3), doesn't the Squeeze theorem establish not just the value, but the existence of the limit as well? – Ragib Zaman Oct 12 '11 at 1:26
In $(3)$, we have chosen a fixed $\epsilon$, which gives us $M$. All we know is that for $x>M$, $\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x$ is somewhere between $\left[1+\frac{a-\epsilon}{x}\right]^x$ and $\left[1+\frac{a+\epsilon}{x}\right]^x$, but it could be anywhere between those two values. The limit of the lower bound is $e^{a-\epsilon}$ and the limit of the upper bound is $e^{a+\epsilon}$, so as $x\to\infty$, $\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x$ could be bouncing around between those values. So $(3)$ only gives a bound on $\liminf$ and $\limsup$. – robjohn Oct 12 '11 at 1:55
@Ragib: However, $\epsilon$ was arbitrarily small, so $(4)$ says that $\liminf=\limsup$, so the limit exists and is equal to them. The Squeeze theorem only holds if the limits of the two bounds are equal, which they are not here. – robjohn Oct 12 '11 at 1:58
Aah thank you @robjohn, I didn't absorb that subtlety even when looking for one! Perhaps you could incorporate your comments into the answer, it may inform other people like me. – Ragib Zaman Oct 12 '11 at 2:01
@Ragib: so incorporated. – robjohn Oct 12 '11 at 2:12