Solving limits by Taylor series

Question

Perhaps I didn't fully understand the concept of Taylor series. I would like to compute

$$\lim_{x \to 1} \frac{\ln x}{x^{2}-1}$$

using Taylor expansion around the right point (point is not given).

So far I only solved 'series problems' when the point was given. I could just use l'Hopital's rule, but that's not the point of the exercise, I guess.

Answer: $0.5$.

Answer 1

You need expansion around the point you want to find limit in. $\ln x = \ln(1 + (x - 1)) = (x - 1) + o(x - 1)$, $x^2 - 1 = (x - 1) (x + 1) = (x - 1) \cdot 2 + o(x - 1)$ (all $o$ with base $x \to 1$). So the expression under limit is $\frac{(x - 1) + o(x - 1)}{2(x - 1) + o(x - 1)} = \frac{1}{2} + o(1)$.

Answer 2

May be helpful to change variables to let $u = x-1 \to 0$ then the limit becomes $$ \begin{split} \lim_{x \to 1} \frac{\ln x}{x^2-1} &= \lim_{x \to 1} \frac{\ln x}{(x+1)(x-1)} \\ &= \lim_{u \to 0} \frac{\ln (1+u)}{u(u+2)} \\ &= \lim_{u \to 0} \frac{u - \frac{u^2}{2}+\frac{u^3}{3} \ldots}{u(u+2)} \\ &= \lim_{u \to 0} \frac{1 - \frac{u}{2}+\frac{u^2}{3} \ldots}{u+2} \\ &= \frac12 \end{split} $$