I have to solve the following limit using Taylor's series:
$$
\lim_{x \to 1} x^{\frac{1}{1-x}}
$$
I can perfectly solve this limit using normal limit properties – such as L'Hospital and chain rules -, but I'm struggling to solve it using Taylor's expansion. I tried the following:
I start with a basic $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$.
Then, I rearrange the original limit and substitute the above result in it:
$$
\begin{align*}
\lim_{x \to 1} x^{\frac{1}{1-x}} &= \lim_{x \to 1} e^{\ln{x^{\frac{1}{1-x}}}}\\
&= \lim_{x \to 1} e^{\frac{1}{1-x}\ln{x}}\\
&= \lim_{x \to 1} e^{\sum_{n=0}^{\infty} x^n \ln{x}}\\
&= \lim_{x \to 1} e^{(1 + x + x^2 + …) \ln{x}}\\
\end{align*}
$$
And then I struggle. In order to this limit result in $\frac{1}{e}$, the exponent $(1 + x + x^2 + …)\cdot\ln{x}$ must equal -1. However I'm unable to understand how can this be possible, since, to me, it should approach 0 as $x$ approaches 1.
What am I getting wrong?
Best Answer
There are two issues in your solution:
(1). For $\frac{1}{1-x}=\sum_{k=1}^\infty x^k$ to be held, $x$ must satisfy $|x|<1$.
(2). When you handle $\lim_{x\to1}(1 + x + x^2 + ...) \ln{x}$, you use $\infty \cdot 0=0$, it is not true.
Therefore you can't use (1). What you have to do is to use the Taylor series of $\ln x=\ln(1+(x-1))$ at $x=1$ and then you will get the answer.