Proving the existence of number such that $\lim_{h\to 0}(x^h-1)/h=1$

Question

I want to prove that there exists a number such that $$\lim_{h\to 0} \frac{x^h-1}{h}=1$$

So to prove it you can say let $$f(x) = \lim_{h\to 0} \frac{x^h-1}{h}$$ Then by interchange of limit and differentiation operator $f'(x)= 1/x$.

My first question is :How to somehow show you can interchange limit and differentiation operator in this case?I know that who can not exchange them always .

Then you can say $$f(x)= \int_1^x\frac1x\, dx $$ Now you need to show this integral increases without bound. It is easy to see it is continuous on the interval $(0,\infty]$ since $1/x$ is continuous on that interval. And so, if it increase without bound, it must take value $1$.

How to show it increases without bound? And is my above proof correct?

Answer 1

Let $$ \alpha_n=\left(1+\frac1n\right)^n\quad\text{and}\quad\beta_n=\left(1+\frac1n\right)^{n+1}\tag1 $$ Then, for non-zero $|h|\le\frac1n$, Bernoulli's Inequality says $$ \begin{align} \frac{\alpha_n^h-1}h &=\frac{\left(1+\frac1n\right)^{nh}-1}h\\[3pt] &\le\frac{(1+h)-1}h\\[6pt] &=1\tag2 \end{align} $$ Note that for $h\lt0$, the sense of the inequality in the numerator changes, but then the negative denominator restores the sense. Furthermore, $$ \begin{align} \frac{\alpha_n^h-1}h &=\frac{\left(1-\frac1{n+1}\right)^{-nh}-1}h\\[3pt] &\ge\frac{\left(1+h\frac{n}{n+1}\right)-1}h\\[6pt] &=\frac{n}{n+1}\tag3 \end{align} $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{\frac{n}{n+1}\le\frac{\alpha_n^h-1}h\le1}\tag4 $$ Similarly, for non-zero $|h|\le\frac1{n+1}$ $$ \begin{align} \frac{\beta_n^h-1}h &=\frac{\left(1+\frac1n\right)^{(n+1)h}-1}h\\[3pt] &\le\frac{\left(1+h\frac{n+1}n\right)-1}h\\[6pt] &=\frac{n+1}n\tag5 \end{align} $$ and $$ \begin{align} \frac{\beta_n^h-1}h &=\frac{\left(1-\frac1{n+1}\right)^{-(n+1)h}-1}h\\[3pt] &\ge\frac{(1+h)-1}h\\[6pt] &=1\tag6 \end{align} $$ Thus, $$ \bbox[5px,border:2px solid #C0A000]{1\le\frac{\beta_n^h-1}h\le\frac{n+1}n}\tag7 $$ Since $\frac{x^h-1}h$ is monotonically increasing in $x$ for non-zero $|h|\lt\frac1{n+1}$, $(4)$ and $(7)$ say that for all $x\in[\alpha_n,\beta_n]$ and non-zero $|h|\lt\frac1{n+1}$, $$ \frac{n}{n+1}\le\frac{x^h-1}h\le\frac{n+1}n\tag8 $$ As is shown in this answer, $\alpha_n$ is increasing and $\beta_n$ is decreasing. Since $[\alpha_n,\beta_n]$ is a decreasing nested sequence of non-empty compact sets, Cantor's Intersection Theorem says that $$ \bigcap_{n=1}^\infty\,[\alpha_n,\beta_n]\ne\emptyset\tag9 $$ In fact, since $\beta_n-\alpha_n=\frac1{n+1}\beta_n$ decreases to $0$, the intersection in $(9)$ consists of one point, called $e$.

Thus, we have that for non-zero $|h|\lt\frac1{n+1}$ $$ \frac{n}{n+1}\le\frac{e^h-1}h\le\frac{n+1}n\tag{10} $$ Inequality $(10)$ and The Squeeze Theorem imply $$ \lim_{h\to0}\frac{e^h-1}{h}=1\tag{11} $$

Answer 2

For $x>0$ fix, let $g(t):=x^t= e^{t \ln x}.$

Then $g'(t)=x^t \cdot \ln x.$ Thus $g'(0)= \ln x.$

On the other hand, $ \lim_{h\to 0} \frac{x^h-1}{h}= \lim_{h \to 0}\frac{g(h)-g(0)}{h-0}=g'(0).$

Hence

$$ \lim_{h\to 0} \frac{x^h-1}{h}=1 \iff \ln x=1 \iff x=e.$$

Consequence: $g(t)=e^t.$

Answer 3

If $\frac {x^h - 1}{h} = 1$ then $x = (1+h)^\frac {1}{h}$

Now taking the limit as $h$ approaches $0,$ will give us the value we seek for $x$

$x = \lim_\limits{h\to 0} (1+h)^\frac 1h = \lim_\limits{n\to \infty} (1+\frac {1}{n})^n = e$

by definition.

Actually, if you only need to prove existence,

$f(x)$ is continuous.

$f(1) = 0$

$f(4) = \lim_\limits{n\to \infty} \frac {4^\frac 1n - 1}{\frac 1n}$

Multiply numerator and denominator by $(4^{1-\frac {1}{n}} + 4^{1-\frac {2}{n}} + \cdots + 4^{1-\frac {n-1}{n}} + 1)$

$f(4) = \lim_\limits{n\to \infty} \frac {4 - 1}{\frac 1n(4^{1-\frac {1}{n}} + 4^{1-\frac {2}{n}} + \cdots + 4^{1-\frac {n-1}{n}} + 1)}$

$4^\frac 12 = 2$

If ${1-\frac {k}{n}} \le \frac 12$ then $4^{1-\frac {k}{n}} < 2$

and if $\frac 12 < {1-\frac {k}{n}} < 1$ then $4^{1-\frac {k}{n}} < 4$

$(4^{1-\frac {1}{n}} + 4^{1-\frac {2}{n}} + \cdots + 4^{1-\frac {n-1}{n}} + 1) < \frac n2\cdot 4 + \frac n2 \cdot 2 = 3n$

$f(4) > \lim_\limits{n\to\infty}\frac {3}{(\frac 1n) 3n} = 1$

By the IVT, there exists $1<x<4$ such that $f(x) = 1$

Answer 4

A non-constructive proof:

Let $$f(x):=\lim_{h\to0}\dfrac{x^h-1}h.$$ We will prove that $f$ is continuous and $f(x)$ straddles $1$.

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By the generalized binomial development, $$f(x+\delta)-f(x)=\lim_{h\to0}\dfrac{(x+\delta)^h-x^h}h \\=\lim_{h\to0}x^h\dfrac{1+h\delta+h(h-1)\dfrac{\delta^2}2+h(h-1)(h-2)\dfrac{\delta^3}{3!}+\cdots-1}h \\=\lim_{h\to0}\left(\delta+(h-1)\frac{\delta^2}2+(h-1)(h-2)\frac{\delta^3}{3!}+\cdots\right)<\delta $$

for $\delta$ is sufficiently small (the sum converges when $|\delta|<1$).

Hence for any $\epsilon$, we can find $\delta$ such that

$$|f(x+\delta)-f(x)|<\epsilon$$ and this proves the continuity of $f$.

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Next, by the binomial theorem, $$2n^n<(n+1)^n=n^n+n^n+(n-1)\frac{n^{n-1}}2+(n-1)(n-2)\frac{n^{n-2}}{3!}+\cdots<n^n\left(1+1+\frac12+\frac1{3!}+\cdots\right)<3n^n$$ (last step because $n!\ge 2^n$).

Hence,

$$2<\left(1+\frac1n\right)^n<3$$

and

$$n(2^{1/n}-1)<1,\\n(3^{1/n}-1)>1.$$

Hence by the IVT, there must be an $x\in(2,3)$ such that $f(x)=1.$

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Final note:

If we want, can refine the value of $x$ such that $f(x)=1$ by a tighter bracketing of

$$\left(\frac{n+1}n\right)^n.$$

An upper bound is

$$1+1+\frac12+\frac1{3!}+\cdots$$

Remains to check if it is tight.