3
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I guess this really seems to be easiest and fastest way. (Though the fractions make it look kinda unintuitive at first) $\endgroup$ – Imago Jan 30 '16 at 10:43
3
$\begingroup$

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 $

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.