# 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?

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 $\lim_{x\rightarrow \infty} e^x = +\infty$ $$\implies \text{ Rqrd. limit} = {-1}$$

• I guess this really seems to be easiest and fastest way. (Though the fractions make it look kinda unintuitive at first) – Imago Jan 30 '16 at 10:43

A slight variation of Tim's answer given above.

Substituting: $x^2-x$ with $\log u$

$\lim\limits_{x \to -\infty} \frac {1-e^{x^2-x}}{1+e^{x^2-x}} =$ $\lim\limits_{\log u\to \infty} \frac {1-e^{\log u}}{1+e^{\log u}} = \lim\limits_{u\to \infty}\frac{1-u}{1+u}$ $= \lim\limits_{u\to \infty} \frac{1}{1+u} + \frac{-u}{1+u} = 0 + (-1) =-1$

Note that $$\frac {1-e^{x^2-x}}{1+e^{x^2-x}}\sim_{-\infty} \frac {-e^{x^2-x}}{+e^{x^2-x}}=-1$$