# Finding $\lim_{n\to \infty} \ln(\frac{n}{n+1})$ using continuity

I have a confusion about finding limits "using the continuity" of a function. I have two concrete examples that I think illustrates my confusion.

The first example is $$\lim_{n\to \infty}\lim\left( \frac{n}{n+1}\right).$$ The solution is to say that because $\ln$ is continuous one can bring the limit inside so: $$\ln\left(\lim_{n\to \infty} \frac{n}{n+1}\right) = \ln(1) = 0.$$ I know the "definition" of continuity that says you can $$\lim_{x\to a} f(x) = f(a)$$ when $f$ is continuous at $a$. My confusion is how this applies to the limit above there the $n$ approaches $\infty$ and not a number $a$.

Another example that I am confused about is $$\lim_{x\to 1^-} \ln\left(\frac{1}{1-x}\right) = \ln\left(\lim_{x\to 1^-}\frac{1}{1-x}\right) = \infty.$$ Again this is because $\ln$ is continuous. From what I understand of a function being continuous just means that it is continuous at every point in the domain. How does this work when the "inside" of the function does to infinifty?

Again, I am not confused about how to actually do this, I am confused about how the definition of continuity let's us do this.

Theorem 1 : You have that $f$ continuous at $a$ $\iff$ for all sequences $(x_n)_n$ s.t. $x_n\to a$, you have $$\lim_{n\to \infty }f(x_n)=f(a)=f\left(\lim_{n\to \infty }x_n\right).$$
Theorem 2 : $\lim_{x\to a}f(x)=\ell$ with $a,\ell\in\mathbb R\cup\{-\infty ,\infty \}$ $\iff$ for all sequence $(x_n)_n$ s.t. $x_n\to a$ you have $$\lim_{x\to a}f(x)=\ell.$$
• $x\longmapsto \ln(x)$ is continuous at $x=1$ and $\left(\frac{n}{n+1}\right)_{n}$ is a sequence that converge to $1$. Using the theorem 1, the claim follow.
For your other example, you know that $$\lim_{x\to \infty }\ln(x)=\infty .$$
• Since $\frac{1}{1-x}\to +\infty$ when $x\to -1^+$, the claim follow with the theorem 2.