I'm currently stuck on a problem and would appreciate any help or guidance that you can provide. The problem asks to use Taylor series expansion to evaluate the following limit:
$$\lim_{x\rightarrow 1} \frac{\ln x-\left( x-1\right) }{\left( x-1\right)^{2} }$$
Here's what I've tried so far:
I started by finding the first few derivatives of $f(x) = \ln x$ evaluated at $x=1$:
$$f(1) = \ln 1 = 0$$
$$f'(1) = \frac{d}{dx} \ln x \bigg|_{x=1} = 1$$
$$f''(1) = \frac{d^2}{dx^2} \ln x \bigg|_{x=1} = -1$$
$$f'''(1) = \frac{d^3}{dx^3} \ln x \bigg|_{x=1} = 2!(-1) = 2$$
Using the Taylor series expansion for $f(x) = \ln x$ about $x = 1$, we get:
$$\ln x = \sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!}(x-1)^n = 0 + 1(x-1)^1 - \frac{1}{2}(x-1)^2 - \frac{1}{3!}(x-1)^3 + \cdots$$
Next, I tried to apply this series to the given limit by replacing $\ln x$ with the series and simplifying:
$$ \begin{aligned} \lim _{x \rightarrow 1} \frac{\ln x-(x-1)}{(x-1)^{2}} &= \\ \lim _{x \rightarrow 1} \frac{\left[1(x-1)-\frac{1}{2}(x-1)^2 - \frac{1}{3!}(x-1)^3 + \cdots\right]-(x-1)}{(x-1)^2} &= \\ \lim _{x \rightarrow 1} \frac{-\frac{1}{2}(x-1) - \frac{1}{3!}(x-1)^2 + \cdots}{(x-1)} &= \\ \lim _{x \rightarrow 1} \frac{-\frac{1}{2} - \frac{1}{3!}(x-1) + \cdots}{1} &= -\frac{1}{2} \end{aligned} $$ However, I'm not sure if my approach is correct or if there's a simpler way to solve this problem using Taylor series expansion. Any feedback or advice would be greatly appreciated. Thank you in advance!