Find for which $a \in \mathbb{R}$ $\frac{e^x(1-\log x)}{x^3-x^2} = o(x^a)$ For $x \to 0^+$.

Question

Find for which $a \in \mathbb{R}$ $\frac{e^x(1-\log x)}{x^3-x^2} = o(x^a)$ For $x \to 0^+$.

I know that $f(x)= o(g(x))$ if $\lim_x\to{x_0} \frac{f(x)}{g(x)} = 0$

So I should find $a$ for which $\lim_{x\to{0^+}} \frac{e^x(1-\log x)}{x^a(x^3-x^2)} = 0$.

1)If $a>0$ I have the indeterminate form $\frac{\infty}{0}$.

2)If $a<0$ I have the indeterminate form $\frac{0 \infty}{0}$.

I tried to apply de l'Hôpital rule but doesn't work.

I think that Taylor Series colud be helpful but I can't see any way to relate the function with standard Taylor series (except for $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$).

Do you have any hint?

Answer 1

Sorry I was unclear (if not wrong) in my comment.

$$\lim_{x\to{0^+}} \dfrac{e^x(1-\log x)}{x^a(x^3-x^2)} = \lim_{x\to{0^+}}\dfrac{e^x}{x-1}\dfrac{1 - \log x}{x^{a+2}}$$

First term converges (to $-1$), so it is irrelevant, and you are left with $\lim_{x\to{0^+}}\dfrac{1 - \log x}{x^{a+2}}$. Now l'Hôpital is your friend.