For calculating a vertical asymptote, I need to evaluate for $x$ going to $-1$:
$f(x) = \ln(x/x+1)$
$\lim_{x\to -1} \ln(x/(x+1)) $
After applying the chain rule for limits, it seems I need to evaluate $\ln(x)$ for $x$ going to minus infinity. That's outside the definition of this function. But, apparently the answer is positive infinity. How to derive limits for $x$ reaching outside the definitions of functions? Or am I applying the chain rule incorrectly?
thx in advance
babi
The function $f(x)=\ln\frac{x}{x+1}$ is defined where $$ \frac{x}{x+1}>0 $$ that is, in the set $(-\infty,-1)\cup(0,\infty)$.
Since $$ \lim_{x\to-1^-}\frac{x}{x+1}=\infty $$ you have $$ \lim_{x\to-1^-}f(x)=\infty $$ because $\lim_{t\to\infty}\ln t=\infty$.
Similarly, $$ \lim_{x\to0^+}f(x)=\infty $$
So you have two vertical asymptotes at $-1$ (from the left) and at $0$ (from the right).