Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$.

Question

Problem

Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.

Solution

Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain

$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$ $$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$ Therefore \begin{align*} \lim_{x \to 1} \frac{x-x^x}{1-x+\ln x}&=\lim_{x \to 1}\frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2)]}\\ &=\lim_{x \to 1}\frac{-(x-1)^2-o((x-1)^2)}{-\frac{1}{2}(x-1)^2+o((x-1)^2)}\\ &=2. \end{align*}

Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

Answer 1

$1-x=h\implies$

$$\lim_{h\to0}(1-h)\cdot\dfrac{1-(1-h)^{-h}}{h+\ln(1-h)}=\lim_{h\to0}\dfrac{1-\left(1+(-h)(-h)+\dfrac{(-h)(-h-1)}2\cdot h^2+O(h^3)\right)}{h-\left(h+\dfrac{h^2}2+O(h^3)\right)}=\dfrac1{\dfrac12}$$

So, you are correct

Answer 2

Why no? $$\lim_{x\rightarrow1}\frac{x-x^x}{1-x+\ln{x}}=\lim_{x\rightarrow1}\frac{1-x^x(1+\ln{x})}{-1+\frac{1}{x}}=\lim_{x\rightarrow1}\frac{-x^x(1+\ln{x})^2-x^x\cdot\frac{1}{x}}{-\frac{1}{x^2}}=2$$

Answer 3

Another solution by L'Hospital's rule

Let $x=:1+h$. Then \begin{align*} \lim_{x \to 1} \frac{x-x^x}{1-x+\ln x}&=\lim_{h \to 0}\frac{(1+h)[1-(1+h)^{h}]}{\ln(1+h)-h}\\ &=\lim_{h \to 0}\frac{e^{h\ln(1+h)}-1}{h-\ln(1+h)}\\ &=\lim_{h \to 0}\frac{h\ln(1+h)}{h-\ln(1+h)}\\ &=\lim_{h \to 0}\frac{h^2}{h-\ln(1+h)}\\ &=\lim_{h \to 0}\frac{2h}{1-\frac{1}{1+h}}\\ &=2\lim_{h \to 0}(1+h)\\ &=2. \end{align*}