Exponential Limits And Indeterminate Forms

Question

Consider two functions $f(x)=x^{\frac{1}{x-1}}$ and $g(x)=(x^2)^{\frac{1}{x-1}}$ Now $$\lim_{x \to 1^+}x=\lim_{x \to 1^+}x^2=1$$ and also $$\lim_{x \to 1^+}\frac{1}{x-1}= \infty$$ but $\lim_{x \to 1^+}f(x)$ is not equal to $\lim_{x \to 1^+}g(x)$ Why????

Answer 1

This is just like saying that $$\lim_{x\to \infty}\left(1+{1\over x}\right)^x=1^\infty=1$$which is wrong and we all know the answer is $e\approx 2.718\cdots$.

Based on a same argument, we have $$\lim _{x\to 1^+}x^{1\over x-1}=\lim _{x\to 0^+}(1+x)^{1\over x}=\lim_{x\to \infty}\left(1+{1\over x}\right)^x=e$$and $$\lim _{x\to 1^+}(x^2)^{1\over x-1}=\lim _{x\to 0^+}(1+x)^{2\over x}=\lim_{x\to \infty}\left(1+{1\over x}\right)^{2x}=e^2$$