solving equation with limit: $1+x =\lim_{n\to\infty}(1+\frac{y}{n})^n => y = \lim_{n\to\infty}n((1+x)^\frac{1}{n} - 1)$

Question

I'm trying to understand a part of logarithmic series expansion proof below:

The logarithmic function is the inverse of the exponential function, so that

$y=ln(1+x)$ implies $1+x=e^y = \lim_{n\to\infty}(1+\frac{y}{n})^n$

Unscrambling the limit to solve for $y$ gives

$y = \lim_{n\to\infty}n((1+x)^\frac{1}{n} - 1)$

... I can't understand the "unscrambling the limit to solve for y" part, i.e.

$1+x = \lim_{n\to\infty}(1+\frac{y}{n})^n => y = \lim_{n\to\infty}n((1+x)^\frac{1}{n} - 1)$

I'd be happy to see an explanation of the solution. Thank you

Answer 1

Your brackets (or the source's, I have no way to know) are wrong. The expression should be $$ \log(1+x)=\lim_{n\to\infty}\,n\big[(1+x)^{1/n}-1\big]. $$ And this is simply, with $z=1+x$ $$ n\big[z^{1/n}-1\big]=n\big[e^{\frac1n\log z}-1\big] =n\,\big[\frac1n\log z+o\big(\frac1{n^2}\big)\big] =\log z+o\big(\frac1n). $$ So in the limit you get $\log z=\log(1+x)$.