TITLE: Limit and ln switch
QUESTION [6 upvotes]: Why is 
$$\lim_{x\to\infty}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)=\ln\left(\lim_{x\to\infty}\frac{1+1/x}{\sqrt{1-1/x+1/x^2}}\right) ?$$  I've seen this way of rewriting, but I can't see why it's equal.

REPLY [3 votes]: We know that if $f(u(x))$ is continuous function, then $\lim_{x\to x_0}f(u(x))=f\left(\lim_{x\to x_0}u(x)\right)$
Eg: $f(u(x))=\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$ is a continuous function!