Prove that $ \ln\left(e +\frac{1}{n}\right) \to 1$ as $n$ approaches $\infty$.
I know I must show $\exists$ $n > N$ such that $\left|\ln\left(e +\frac{1}{n}\right)-1\right|< \varepsilon $
But I am having difficulty simplifying the expression and isolating for n. Is there an elementary log rule that can be used?
On
This is a good result/corollary to know about continuity: If $f$ is continuous at $x$, then: $$\lim_{n \rightarrow \infty} x_n = x \rightarrow \lim_{n \rightarrow \infty}f(x_n) = f(x)$$ In this case, the function $\ln$ is continuous at $e$ (and in an open set about $e$) , so that, applying that to $f(x)=\ln x $, we have that : $$\lim_{n \rightarrow \infty} \ln(e+1/n)= \ln( e+ \lim_{n \rightarrow \infty} 1/n)=\ln e $$
Edit: @Did suggested this corollary was written ambiguously, or in a mangled way: the intended point is that continuity implies sequential * continuity, which is to say that , when f is continuous at a point x, then, whenever we have $x_n \rightarrow x$, it follows that $f(x_n) \rightarrow f(x)$.
Formally, here, we have :i) ln(x) is continuous in $( 0, \infty)$ ii) ln(x) is sequentially continuous in $(0, \infty)$ . Then we have $e+1/n \rightarrow e$, from which we conclude, using sequential continuity, that $ln(e+1/n) \rightarrow ln(e)=1$
Hint: Rewrite as $\ln\left(e\left(1+\frac{1}{en}\right)\right)$, and then as $\ln e+\ln\left(1+\frac{1}{en}\right)$. Then use the fact that for $x\gt 0$ we have $0\lt \ln(1+x)\lt x$.