Solve $\lim_{x \to 1} x^{\frac{1}{1-x}}$ using Taylor's expansion

Question

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?

Answer 1

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.

Answer 2

I'll take it from your last line. We have $$L=\lim _{x\rightarrow 1}\exp\left(\sum ^{\infty }_{k=0} x^{k} \cdot \ln( x)\right)$$ Then using a first order Taylor expansion of $\ln$ about $x=1$, $$L \approx \lim _{x\rightarrow 1}\exp\left(\sum ^{\infty }_{k=0} x^{k} \cdot ( 1-x)\right)$$ $$=\lim _{x\rightarrow 1}\frac{\exp\left(\sum ^{\infty }_{k=0} x^{k}\right)}{\exp\left(\sum ^{\infty }_{k=0} x^{k+1}\right)}$$ $$=\lim _{n\rightarrow \infty }\lim _{x\rightarrow 1}\frac{\prod ^{n}_{k=0}\exp\left( x^{k}\right)}{\prod ^{n}_{k=0}\exp\left( x^{k+1}\right)}$$ $$=\lim _{n\rightarrow \infty }\lim _{x\rightarrow 1}\frac{1}{\exp\left( x^{n+1}\right)}$$ $$=\lim _{n\rightarrow \infty }\frac{1}{e} =\frac{1}{e}.$$

Answer 3

You're making things more complex than they really are. By the substitution $x=1+h\;(h\to 0)$, you only need an expansion at order $1$, i.e. a standard high school limit. Indeed $$ x^{\tfrac{1}{1-x}}=\mathrm e^{\tfrac{\ln x}{1-x}}=\mathrm e^{-\tfrac{\ln (1+h)}{h}}$$ Now , it is well-known from high school that $\lim\limits_{h\to 0}\dfrac{\ln (1+h)}{h}=1$, whence the sought limit $\mathrm e^{-1}$.