Sum of positive infinity and negative infinity

Question

Consider the following function of $\tau$:

$$ h(\tau) := C_1 \ln\left(1-\frac{a}{\tau}\right) - C_2 \ln\left(1-\frac{b}{\tau}\right), $$

where $a > b>0$ and $C_2=\ln(1-b)<C_1=\ln(1-a)<0$. Then it is claimed that

$$ \lim_{\tau \to 0}h(\tau) = \infty. $$

I do not understand why this is the case. As $\tau$ goes to $0$ from the left, we would have $-\infty+\infty$. Why is this equal to $\infty$, please?

Answer 1

\begin{align} h(\tau) := C_1 \ln\left(1-\frac{a}{\tau}\right) - C_2 \ln\left(1-\frac{b}{\tau}\right)= \ln\frac{\left(1-\frac{a}{\tau}\right)^{C_1}}{\left(1-\frac{b}{\tau}\right)^{C_2}}= \ln\frac{\left(\tau-{a}\right)^{C_1}}{\left(\tau-{b}\right)^{C_2}}\tau^{C_2-C_1} \\= \ln\frac{\left(\tau-{a}\right)^{C_1}}{\left(\tau-{b}\right)^{C_2}} + ({C_2-C_1})\ln\tau \end{align}

can you take the limit now?