calculus - Limit problem $e^x$ without L'Hôpital's rule

$$\lim_{x \to -\infty} \frac {1-e^{x^2-x}}{1+e^{x^2-x}}$$
I solved this limit problem by applying L'Hôpital's rule and I got $-1$.

Question: how to solve this limit without L'Hopital rule and Taylor series?

Answer

hint:

Divide Numerator and Denominator by $e^{x^2-x}$

$$\lim_{x \to -\infty} \frac {1-e^{x^2-x}}{1+e^{x^2-x}} = \lim_{x \to -\infty} \frac {\dfrac{1}{e^{x^2-x}}-1}{\dfrac{1}{e^{x^2-x}}+1}$$

Now, $\lim_{x\rightarrow -\infty} \dfrac{1}{e^{x^2-x}} = 0$, since $\lim_{x\rightarrow -\infty} x^2 -x = +\infty$ and $\lim_{x\rightarrow \infty} e^x = +\infty$
$$\implies \text{ Rqrd. limit} = {-1}$$