Prove that $n(x^{1/n}-1)$ converges to $\ln(x)$. I would like to prove that $n(x^{1/n}-1) \rightarrow \ln(x)$ using the epsilon-N definition of convergence, or possibly the pinching theorem. I cannot use derivatives. I need a hint on how to get started.

I assume we know that $1+x\le e^x$. Taking logs yields $\ln(1+x)\le x$, and so also $$ \ln(1+x) = -\ln(\tfrac1{1+x}) = -\ln(1-\tfrac{x}{1+x}) \ge -(-\tfrac{x}{1+x}) = \tfrac{x}{1+x} $$ Thus $$ \frac{x}{1+x} \le \ln(1+x) \le x $$ Replacing $x$ with $x^{1/n}-1$ yields $$ \frac{x^{1/n}-1}{x^{1/n}} \le \ln(x^{1/n}) \le x^{1/n}-1 $$ Thus $$ \ln(x) \le n(x^{1/n}-1) \le x^{1/n} \ln(x) $$ which yields the desired limit by squeezing.