3
$\begingroup$

I'm currently studying limits with $x$ in the exponent. The following formula simplifies the work to solve limits.

If $\lim\limits_{x \to a}{\phi(x)} = 1$ and $\lim\limits_{x \to a}{\psi(x)} = \infty$, then

$\lim\limits_{x \to a}{[\phi(x)]^{\psi(x)}} =$ $\lim\limits_{x \to a}{\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\}^{\alpha(x)\psi(x)}} =$ $e^{\lim\limits_{x \to a}{[\phi(x) - 1]\psi(x)}}$

For the most part, I understand how the formula is derived. However, there's one part I don't understand.

Why does $\lim\limits_{x \to a}{\{[1+\alpha(x)]^{\frac{1}{\alpha(x)}}\}} = e$?

  • 0
    You need to mention the conditions that $\lim_{x\to a} \phi(x) =1$ and $\alpha(x) =\phi(x) - 1\to 0$ as $x\to a$. Then it is a standard result that $(1+\alpha (x)) ^{1/\alpha (x)} \to e$. – 2017-02-16
  • 0
    @ParamanandSingh -- Ah, yes, I forgot about the conditions. However, since I'm new to $e$, the result you mention is not standard to me. :) – 2017-02-16
  • 2
    If you are new to $e$ then you should assume certain facts about $e$ without proof. One of these is the definition of $e$ and there are many ways to define $e$. The simplest approach is to define $e=\lim_{n\to\infty} (1+(1/n))^{n}$ where $n$ is a positive integer. Using this result it can be established that $e=\lim_{x\to 0}(1+x)^{1/x}$ where $x$ is a real variable. For a beginner understanding the proofs for these results is a challenge and one should just accept it on faith. After learning a fair amount of calculus one can revisit the proofs. – 2017-02-16
  • 0
    @ParamanandSingh -- OK. I hesitate to use such approach, but I suppose I must move on. :) Thanks. – 2017-02-16
  • 1
    In case you have already studied a fair amount of calculus then you can have a look at my blog posts which develop a full theory of symbols $e^{x}, a^{x}, \log x$ (all these symbols are connected in some ways and most textbooks of calculus don't define these symbols and ask the students to accept their properties on faith) . The starting post is available at http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-1.html – 2017-02-16
  • 0
    Paramanand's blogs are exceptionally well written and provide depth to a host of topics. – 2017-02-16
  • 0
    Thanks @Dr.MV. Such words are a biggg motivation!! – 2017-02-16
  • 0
    Paramnand, you're quite welcome. Aside from being a friend here, I've read several of your blogs and objectively attest to their quality and thoroughness. It's work with which to be proud. -Mark – 2017-02-16

1 Answers 1

3

METHODOLOGY $1$: Pre-Calculus Approach

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

Letting $x$ be replaced with $1+\alpha(x)$ in $(1)$ reveals

$$\frac{\alpha(x)}{1+\alpha(x)}\le \log(1+\alpha(x))\le \alpha(x)\tag 2$$

Then, noting that $(1+\alpha(x))^{1/\alpha(x)}=e^{\frac{1}{\alpha(x)}\log(1+\alpha(x))}$ and applying $(2)$, we find that

$$e^{\frac{1}{1+\alpha(x)}}\le (1+\alpha(x))^{1/\alpha(x)}\le e \tag3$$

whence applying the squeeze theorem to $(3)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to a}(1+\alpha(x))^{1/\alpha(x)}=e}$$

METHODOLOGY $2$: Asymptotic Analysis

We have for $\alpha(x) \to 0$ as $x\to a$

$$\begin{align} (1+\alpha(x))^{1/\alpha(x)}&=e^{\frac{1}{\alpha(x)}\color{blue}{\log(1+\alpha(x))}} \\\\ &=e^{\frac{1}{\alpha(x)}\color{blue}{\left(\alpha(x)+O\left(\alpha^2(x)\right)\right))}}\\\\ &=e^{1+O(\alpha(x))}\\\\ &\to e\,\,\text{as}\,\,x\to a \end{align}$$

In the development herein, we used the asymptotic expansion $\displaystyle \log(1+t)=t+O(t^2)$.

  • 0
    Is $\log$ base $e$ (as in $\ln$)? – 2017-02-16
  • 0
    @SirJony Yes, indeed that is the standard notation. – 2017-02-16
  • 0
    Oh, I was taught that the implicit $\log$ base was $10$, and $\ln$ was $e$. Also, I'm not familiar with the second blue part. What is $O$? Is this usually taught by the time students first start solving limits with $x$ in exponents? – 2017-02-16
  • 0
    The notation is called the [Big O Notation](https://en.wikipedia.org/wiki/Big_O_notation#Formal_definition). – 2017-02-16
  • 0
    Are you familiar with Taylor's Theorem or Taylor Series? – 2017-02-16
  • 0
    No, not yet. I'm a beginner in calculus and analysis. Still working on limits. – 2017-02-16
  • 0
    I've added a new section that relies on pre-calculus tools only. – 2017-02-16
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53772/discussion-between-sir-jony-and-dr-mv). – 2017-02-16
  • 0
    Oh yes my comment was delayed for sometime. I am a bit lazy on smartphone. – 2017-02-16
  • 0
    I'm unsure of how you got from (2) to (3). Did you make the 3 expressions the powers of $e$? If so, I got... $$1/(1+\alpha) \leq \ln(1+ \alpha) \leq \alpha \implies$$ $$e^{1/(1+\alpha)} \leq e^{\ln(1+ \alpha)} \leq e^{\alpha} \implies$$ $$e^{1/(1+\alpha)} \leq 1 + \alpha \leq e^{\alpha} \implies$$ $$e^{1/\alpha(1+\alpha)} \leq (1 + \alpha)^{1/\alpha} \leq e$$ – 2017-02-16
  • 0
    @SirJony You missed two important things. First, we have $\frac{\alpha}{1+\alpha}\le \log(1+\alpha)\le \alpha$. Second, we have $$e^{\frac{1}{\alpha}\log(1+\alpha)}\le e^{\alpha/\alpha}=e^1=e$$and $$e^{\frac{1}{\alpha}\log(1+\alpha)}\ge e^{\frac{1}{\alpha}\left(\frac{\alpha}{1+\alpha}\right)}=e^{\frac{1}{1+\alpha}}$$ – 2017-02-16
  • 0
    Alright, thank you very much! I apologize for taking quite a bit of time. – 2017-02-16
  • 0
    You're welcome. My pleasure! Pleased to see that you have it now. Well done. -Mark – 2017-02-16
  • 0
    @Dr.MV Excellent answer, really. I didn't come by this one before, just now that it was linked in another question. Great. +1 – 2017-03-20
  • 0
    @donantonio Thank you my friend! Just curious, but what other answer is linked to this one? – 2017-03-20
  • 0
    @Dr.MV Well Mark: I just can't find it now...It was about the limit of some kind of indeterminate form under certain conditions, and somebody there linked this question...:) – 2017-03-20