Show that $f_n(x) = n(x^{1/n}-1)$ converges pointwise to $f(x) = \ln x$. Let $f_n:(0,\infty) \rightarrow \mathbb{R}$ be defined by $f(x) = n(x^{1/n}-1)$. Show that $f_n$ converges pointwise to $f(x) = \ln x$.
I'm not getting anywhere here. I can't seem to find a way to rewrite $d(f_n(x), f(x)) = |n(x^{1/n}-1)-\ln x| < \epsilon$, such that for any given $\epsilon$, the inequality makes sense whenever $n\geq N$.
What's really throwing me off is that I have no feeling for what the function $f_n(x) = n(x^{1/n}-1)$ is and how I can relate it to $\ln x$, in a sensible way. Arrriving at
$|n \ln \frac{e^{x^{\frac{1}{n}}-1}}{x^{\frac{1}{n}}}| < \epsilon$,
and then examining the limit is the best approach I can think of. If the RHS vanishes as $n\rightarrow \infty$, then certainly there exists a large enough $N$ so that the inequality makes sense. But I don't really get anywhere doing so. Maybe I'm going about this wrong. Any help on the way is very welcome!
 A: A simple option based on recognizing a derivative: 
For $x > 0$, you have $x^{\frac{1}{n}} = e^{\frac{1}{n}\ln x}$. Fix any $x_0 > 0$, and consider the function $f\colon \mathbb{R} \to \mathbb{R}$ defined by $f(t) = e^{t\ln x_0}$. It is differentiable, and $f^\prime(t) = \ln x_0\cdot e^{t\ln x_0}$; and further
$$
f^\prime(t) = \lim_{h\to0} \frac{f(t+h)-f(t)}{h}.
$$
Now, what is $f^\prime(0)=\lim_{h\to0} \frac{f(h)-f(0)}{h} = \lim_{h\to0} \frac{f(h)-1}{h}$? (To conclude, take "$h=\frac{1}{n}$.")
A: Apply the inequality $(x^{1/n}-1)/x^{1/n} < \ln x^{1/n} < x^{1/n}-1$ and the squeeze theorem.
Rearranging we get
$$\ln x< n(x^{1/n} -1) < \ln x \cdot x^{1/n}.$$
A: Fix $x>0.$ Let $g(y)=x^y, y \in \mathbb R.$ Then $g'(y) = x^y\ln x.$ Thus
$$n(x^{1/n} - 1)=\frac{g(1/n)-g(0)}{1/n-0} \to g'(0) = x^0\ln x = \ln x.$$
A: A good way to start:
$$f_n(x) = n(x^{1/n}-1)=\int_1^x u^{\frac{1}{n}-1}du$$
Then, using $\ln x = \int_1^x u^{-1}du$, we have:
$$f_n(x)-\ln x = \int_1^x \left[u^{\frac{1}{n}}-1\right]\frac{du}{u}$$
Can you see how you might bound their difference from here?
A: METHODOLOGY $1$:
Writing $x=1+y$ and applying the binomial theorem, we have for $-1<y\le 1$
$$\begin{align}
\lim_{n\to \infty} n(x^{1/n}-1)&=\lim_{n\to \infty}n\left((1+y)^{1/n}-1\right)\\\\
&= \lim_{n\to \infty}\sum_{m=1}^{\infty}\frac{\prod_{k=1}^{m-1}\left(\frac1n -k\right)}{m!}y^m\\\\
&=\sum_{m=1}^{\infty}\frac{(-1)^{m-1}y^m}{m}\\\\
&=\log(1+y)\\\\
&=\log(x)
\end{align}$$
as was to be shown!
The case for $y>1$ is left as an exercise for the reader.

METHODOLOGY $2$:
Write the limit as
$$\lim_{n\to \infty}n(x^{1/n}-1)=\lim_{n\to \infty} n\left(e^{\frac1n \log(x)}-1\right)$$
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities 
$$1+x\le e^x\le \frac{1}{1-x}$$
for $x<1$.  Therefore, we have 
$$\log(x) \le n\left(e^{\frac1n \log(x)}-1\right)\le \frac{ \log(x)}{1-\frac1n \log(x)}$$
whereupon applying the squeeze theorem produces the coveted limit
$$\lim_{n\to \infty}n\left(e^{\frac1n \log(x)}-1\right)=\log(x)$$
