Calculating using Taylor's series with Remainder: $ \lim \limits_{x \to 1} \frac{\ln x}{x^2+x-2} $

Question

How to Calculate with Taylor's series with Remainder: $$ \lim \limits_{x \to 1} \frac{\ln x}{x^2+x-2} $$

without using L'Hopital's Rule?

Here is what I reached: $$\lim \limits_{x \to 1} \frac{(x-1) + \frac{(x-1)^2 }{2!}+R_2(x)}{x^2+x-2}$$ and I know that $\lim \limits_{x \to 1} \frac{R_n(x)}{(x-1)^n} = 0 $ (and here I'm supposed to insert $n=2$ and use it somehow to find the limit).
But couldn't somehow get a result. I think I'm missing something essential here. can someone tell me if all the things I wrote are true until here and try to complete the solution in a clear way ?

Answer 1

Notice that $ x^2 + x -2 = (x-1)(x+2)$ then

$$\lim_{x \to 1} \frac{(x -1) - \frac{1}{2}(x - 1)^2 + O((x-1)^3)}{x^2 + x -2} = \lim_{x \to 1} \frac{1 - \frac{1}{2}(x -1) + O((x-1)^2)}{(x+2)} = \frac{1}{3}$$

Answer 2

You didn't need to expand the $\ln$ up to $R_2$. It will be sufficient to expand up to $R_1$. For

$$\frac{\ln x}{x^2 + x - 2} = \frac{(x - 1) + o(x - 1)}{(x - 1)(x + 2)} = \frac{1 + o(1)}{x + 2} \to \frac{1}{3}\quad \text{as $x\to 1$}.$$

Answer 3

Continuing from where you left off, $$\lim_{x\to 1} \frac{(x-1) + \frac{(x-1)^2}{2} + R_2(x)}{(x-1)(x+2)} = \lim_{x\to 1} \frac{1}{(x+2)} + \frac{(x-1)}{2(x+2)} + \frac{R_2(x)}{(x-1)(x+2)}$$

The limit of the first term is $1/3$, the second limit is $0$.

As for $\frac{R_2(x)}{(x-1)(x+2)}$ we know that $\frac{R_2(x)}{(x-1)^2} \to 0$ and so then $$\frac{R_2(x)}{(x-1)} = (x-1) \cdot \frac{R_2(x)}{(x-1)^2} \to 0\cdot 0 = 0.$$

This means that the limit of the third term is $0$ as well. Thus the limit is $1/3.$