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.