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

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Fine Man
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  • 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 $(1+\alpha (x)) ^{1/\alpha (x)} \to e$. – Paramanand Singh Feb 16 '17 at 19:36
  • @ParamanandSingh -- Ah, yes, I forgot about the conditions. However, since I'm new to $e$, the result you mention is not standard to me. :) – Fine Man Feb 16 '17 at 19:40
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    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. – Paramanand Singh Feb 16 '17 at 19:56
  • @ParamanandSingh -- OK. I hesitate to use such approach, but I suppose I must move on. :) Thanks. – Fine Man Feb 16 '17 at 20:01
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    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 – Paramanand Singh Feb 16 '17 at 20:01
  • Paramanand's blogs are exceptionally well written and provide depth to a host of topics. – Mark Viola Feb 16 '17 at 20:08
  • Thanks @Dr.MV. Such words are a biggg motivation!! – Paramanand Singh Feb 16 '17 at 21:33
  • 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 – Mark Viola Feb 16 '17 at 21:36

1 Answers1

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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 $\displaystyle \log(1+t)=t+O(t^2)$.

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