Pointwise Convergence of a Nonlinear Sequence of Functions Let $f_n\left(x\right)=n\left(\sqrt[n]{x}-1\right)$. How can I show that this function converges pointwise to $f\left(x\right)=\ln x$ on $\left(0,\infty\right)$? This is what I tried to do:
$$\lim_{n\to\infty}f_n\left(x\right)=\lim_{n\to\infty}n\left(\sqrt[n]{x}-1\right)=\lim_{n\to\infty}n\sqrt[n]{x}-\lim_{n\to\infty}n.$$
I know that both terms approach infinity because the radical approaches $1$. However, I have no clue how to make the jump from that last expression to a logarithm. Is this a definition problem, or can I reach that conclusion algebraically? Thanks!
Edit 1: I have been fooling around with the equalities $\ln e^x=x$ and $e^{\ln x}=x$. I am still stuck, nevertheless. This is what I tried:
$$n\left(\sqrt[n]{x}-1\right)=e^{\ln n\left(\sqrt[n]{x}-1\right)}=e^{\ln n+\ln\left(\sqrt[n]{x}-1\right)}.$$I cannot seem to progress further.
 A: You could use L'Hôpital's rule. For $x\ne1$:
$$
\lim_{n\rightarrow\infty} \bigl[ n (\root n\of x-1)\bigr]=
\lim_{n\rightarrow\infty} {x^{1/n}-1\over 1/n}=
\lim_{n\rightarrow\infty} {\ln x \cdot{-1\over n^2} \cdot x^{1/n} \over -1/n^2}=
\lim_{n\rightarrow\infty} \bigl[{\ln x\cdot    x^{1/n}  }\bigr]=\ln x
$$
Of course, for $x=1$, the limit is $0=\ln 1$.
Alternatively, you could write $$
\lim_{n\rightarrow\infty} {x^{1/n}-1\over 1/n}
=\lim_{h\rightarrow 0^+} {x^{h}-1\over h}$$
 Then the limit is recognized as being equal to the derivative of $x^t$ with respect to $t$  evaluated at $t=0$.
A: If $x=1$, then $f_n(1)=0=\ln 1$ for each $n\in \mathbb{N}$.
Now assume $x\neq 1$. Since $x>0$ you have $x=e^{\ln x}$ and $\ln x\neq 0$, thus:
$$\lim_{n\to \infty} n\ (x^{1/n}-1) = \lim_{n\to \infty} \frac{e^{(\ln x)/n}-1}{1/n} =\lim_{n\to \infty} \ln x\ \frac{e^{(\ln x)/n}-1}{(\ln x)/n}=\ln x\; ,$$
because of the fundamental limit $\displaystyle \lim_{y\to 0} \frac{e^y -1}{y} =1$.
Therefore $\displaystyle \lim_{n\to \infty} f_n(x)=\ln x$ for $x\in ]0,\infty[$.
