Can you justify interchange of limit and derivative/integral in this example? Is there any way to prove that $$\large f(x)=\lim_{h\to0} {\frac{x^h-1}h}=\int_1^x1/t\ dt$$ Without knowing that both are the logarithm  It is clear that $f(1)=0$  And we would be done if we can prove that $f’(x)=\frac1x$, but that can only be done if we can switch the limit and differentiation operator and that can not be done always and I could not find any justification for doing so. So can we do so in this case and if so what is the justification?I know this can be proved in a much longer way but if we can justify the interchange the proof will be much shorter .
 A: Paraphrasing this comment
Note that
$$
\int_1^xt^{h-1}\,\mathrm{d}t=\frac{x^h-1}h
$$
For $x\gt1$, the integrand converges uniformly to $\frac1t$ on $[1,x]$. Therefore, the integral of the limit is the limit of the integral.

Uniform Convergence of $\boldsymbol{\lim\limits_{h\to0}{t^{h-1}=t^{-1}}}$
Assume that $t\ge1$ and $|h|\le1$.
$$
\begin{align}
0
&\le\frac{t^{h}-1}{ht}\tag1\\
&=\frac{(1+(t-1))^h-1}{ht}\tag2\\
&\le\frac{(1+h(t-1))-1}{ht}\tag3\\
&=\frac{t-1}t\tag4\\[6pt]
&\le1\tag5
\end{align}
$$
Explanation:
$(1)$: $t\ge1$; if $h\ge0$, $t^h\ge1$; if $h\le0$, $t^h\le1$
$(2)$: algebra
$(3)$: Bernoulli's Inequality; the sense of the inequality
$\phantom{\text{(3):}}$ in the numerator is reversed between $h\in[0,1]$
$\phantom{\text{(3):}}$ and $h\le0$, but there is an $h$ in the denominator
$(4)$: algebra
$(5)$: $t\ge1$
Thus, we have shown that for $t\ge1$ and $|h|\le1$,
$$
\left|\,t^{h-1}-t^{-1}\,\right|\le|h|\tag6
$$
which gives uniform convergence as $h\to0$.

Compact Subsets of $\boldsymbol{(0,1]}$
Given $t\in[\epsilon,1]$ for $\epsilon\gt0$,
$$
\begin{align}
|h|
&\ge\left|\,t^{1-h}-t\,\right|\tag7\\
&=\left|\,t^{h-1}-t^{-1}\,\right|\ t^{2-h}\tag8\\
&\ge\left|\,t^{h-1}-t^{-1}\,\right|\epsilon^3\tag9
\end{align}
$$
Explanation:
$(7)$: apply $(6)$ to $1/t$
$(8)$: pull $t^{2-h}$ out of the difference
$(9)$: since $t\ge\epsilon$ and $|h|\le1$, $t^{2-h}\ge\epsilon^3$
Thus, we have
$$
\left|\,t^{h-1}-t^{-1}\,\right|\le|h|\epsilon^{-3}\tag{10}
$$
which gives uniform convergence as $h\to0$.
A: Here is a much simpler approach which avoids differentiation altogether.
Let's use different symbols for different forms $$f(x) =\lim_{h\to 0}\frac{x^h-1}{h},g(x)=\int_{1}^{x}\frac{dt}{t}\tag {1}$$ The definition of $g$ is easier to handle analytically because integrand $1/t$ is continuous on $(0,\infty) $ and hence $g$ is well defined on $(0,\infty) $.
It can be proved with some effort that the limit used in definition of $f$ exists for all $x>0$. Now using this fact we do a substitution $h=1/n$ where $n$ is a positive integer. This gives us $$f(x) =\lim_{n\to \infty} n(x^{1/n}-1)\tag{2}$$ It is now easy to prove that $f(x) =g(x) $ for all $x>0$. It should be obvious that $f(1)=g(1)=0$ and further it is easily proved that $$f(1/x)=-f(x),g(1/x)=-g(x)\tag{3}$$ so it is sufficient to show that $f(x) =g(x) $ for $x>1$.
Let us choose a partition $$P=\{x_0,x_1,x_2,\dots,x_n\} $$ of $[1,x]$ such that $x_k=q^{k} $ where $q^n=x$ and we choose tag points $t_k=x_{k-1}$. The corresponding Riemann sum for the integral defining $g(x) $ is $$\sum_{k=1}^{n}\frac{x_{k}-x_{k-1}}{x_{k-1}}=\sum_{k=1}^{n} \frac{q^k-q^{k-1}}{q^{k-1}}=n(q-1)=n(x^{1/n}-1)$$ and thus the integral equals the limit of this Riemann sum and we get $$g(x) =\lim_{n\to \infty} n(x^{1/n}-1)=f(x)$$

Here is a proof based on discussion in comments that the limit in question exists for all $x>0$. For this we let $x>0$ be fixed and consider $F(h)=x^h$. If $x>1$ then $F(h) >1$ if $h>0$ and $F(h) <1$ if $h<0$. These inequalities get reversed if $0<x<1$. Since $$F(t+s) =F(t) F(s) $$ it follows that for $x>1$ the function $F$ is strictly increasing and for $0<x<1$ it is strictly decreasing (for $x=1$ it remains a constant).
Thus $F(h) $ is a monotone function of $h$. It follows via a standard theorem on monotone functions that $F$ is continuous everywhere except at most a countable number of points. Thus $F$ is continuous at some point $a$. And we have $$F(h) =F(h-b+a+b-a) =F(h-b+a) F(b-a) $$ If $h\to b$ then $h-b+a\to a$ and thus by continuity at $a$ we have $F(h-b+a) \to F(a) $ and so $F(h) \to F(a) F(b-a) =F(b) $ as $h\to b$. This proves that $F$ is continuous at any point $b$ and so it is continuous everywhere.
It follows that $$G(t) =\int_{0}^{t}F(h)\,dh$$ exists and $G'(h) =F(h) $ for all $h$. Integrating the functional equation $$F(t+h) =F(t) F(h) $$ with respect to $h$ we get $$G(t+h) - G(t) =F(t) G(h) $$ Note that $G(0)=0$ and if $G$ is a constant then $F=G'$ is also a constant. Otherwise there is an $h$ such that $G(h) \neq 0$. And then we have $$F(t) =\frac{G(t+h) - G(t)} {G(h)} $$ and the right hand side is clearly a differentiable function of $t$ so that $F$ is differentiable everywhere with derivative $$F'(t) =\frac{F(t+h) - F(t)} {G(h)} $$ In particular $F'(0)$ exists and this means that the limit in question exists.
A: Note that $\int_1^x t^{h-1} \mathrm{d} t = \frac{t^h}{h}\big\vert_1^x = \frac{x^h - 1}{h}$. It suffices to prove that
$$\lim_{h\to 0} \int_1^x \frac{t^h - 1}{t} \mathrm{d} t = 0.$$
We split into three cases:

*

*$x > 1$: Note that, for $1 \le t \le x$,
$$0 \le \frac{t^h - 1}{t} \le x^h - 1$$
and
$$0 \le \int_1^x \frac{t^h - 1}{t}\mathrm{d} t \le (x^h - 1)(x-1).$$
Note that $\lim_{h\to 0} (x^h - 1)(x-1) = 0$. Thus, by the squeeze theorem, we have $\lim_{h\to 0} \int_1^x \frac{t^h - 1}{t} \mathrm{d} t = 0$.


*$0 < x < 1$: We have $\int_1^x \frac{t^h - 1}{t} \mathrm{d} t = \int_x^1 \frac{1 - t^h}{t} \mathrm{d} t$.
Note that, for $x\le t \le 1$,
$$0 \le \frac{1 - t^h}{t} \le \frac{1 - x^h}{x}$$
and
$$0 \le \int_x^1 \frac{1 - t^h}{t} \mathrm{d} t \le \frac{1 - x^h}{x}(1-x).$$
Note that $\lim_{h\to 0} \frac{1 - x^h}{x}(1-x) = 0$. Thus, by the squeeze theorem, we have $\lim_{h\to 0} \int_x^1 \frac{1 - t^h}{t} \mathrm{d} t = 0$.


*$x=1$: It is obvious.
We are done.
