# Investigating the behaviour of a function around the asymptotes

I am interested in sketching the function with the following equation:

$$f(x) = \frac{e^{ax}}{1-x^2},$$ where a is positive.

1. There is no symmetry

2. There are asymptotes at $x = 1$ and $x = -1$

3. The function does not cross the $x$-axis and $f(0) = 1$

4. As $x$ tends to positive infinity, the function tends to $-\infty$; as $x$ tends to $-\infty$, the function tends to $0$.

5. Now, the problematic section: I wanted to look at how the function behaves around the asymptotes and so I first considered $x=1$. I let $x=u+1$ as $u$ tends to zero (by subbing new $x$ in terms of $u$ into the function): $$f(x) = \frac{e^{a(u+1)}}{1-(u+1)^2} = - \frac{e^{a(u+1)}}{u^2 + 2u}$$

as u tends to $0-$, $f(x)$ should tend to zero as a negative sign will flip the exponential (in reality the function is tending to minus infinity)

as u tends to $0+$, $f(x)$ should tend to minus infinity due to the negative sign in front of the function that I obtained (in reality it tends to positive infinity)

A similar process for $x=-1$ by letting my new $x$ to be $x=u-1$ as $u$ tends to $0$: $$f(x) = \frac{e^{a(u-1)}}{-u^2 + 2u}$$

as $u$ tends to $0-$, $f(x)$ should tend to $0$ as the exponential will be flipped because of negative sign (in reality it tends to positive infinity)

as $u$ tends to $0+$, $f(x)$ will tend to negative infinity as in denominator $u^2>2u$ and so the negative sign in front of $u^2$ will yield in negativity. (in reality it does tend to minus infinity)

6. Finally, I did find stationary point and there is a maximum and a minimum. Here they are just in case: $x= \frac{1 \pm \sqrt{1+a^2}}{a}$

But this is not that important as I had problems in finding the behavior of the function around $x=1$ and $x=-1$ (and as you can see I got them wrong). I hope what I wrote makes sense and I would appreciate any advice. Thanks! Please note: this seems as a very laborious method but I was taught this way :)

This is very simplistic, but agrees (mostly) with Claude Leibovici's answer.

For $x\approx1$:

$$f(x)\approx\frac{e^a}{1-x^2}=\frac{e^a}2\left(\boxed{\frac1{1-x}}+\frac1{1+x}\right)$$

For $x\approx-1$:

$$f(x)\approx\frac{e^{-a}}{1-x^2}=\frac{e^{-a}}2\left(\frac1{1-x}+\boxed{\frac1{1+x}}\right)$$

The boxed portion is the main behavior of the function, and when multiplied by the coefficient in front, agrees with the first term of Leibovici's expansion.

• (+1) Very clever! However, it should be completed by the explicite discussion of the cases $x \to 1^+$ and $x \to 1^-$ resp. $x \to -1^+$ and $x \to -1^-$ Oct 5, 2016 at 7:03
• Hi, thank you for your post. I was just wondering should not be there a negative sign before the brackets since when simplifying it we get e^(a)*(1/x^2 -1) and not {1-x^2} in the denominator? Oct 6, 2016 at 8:09
• @Blondie What do you mean by simplifying? I started straight from the original problem $f(x)=\frac{e^{ax}}{1-x^2}$. Oct 6, 2016 at 13:10
• @SimpleArt what I am saying is that (1/x-1) - (1/x+1) gives 2/x^2-1 whilst in the initial problem in the denominator we have 1-x^2. Thus, we need a negative sign in front of the bracket. Oct 6, 2016 at 22:47
• @Blondie Oh... that is most definitely my bad. Oct 7, 2016 at 0:29

May be, you could consider using Taylor series.

Around $x=1$ $$e^{ax}=e^a+a e^a (x-1)+O\left((x-1)^2\right)$$ which makes $$y=\frac{e^{ax}}{1-x^2}=-\frac{e^{ax}}{(x-1)(x+1)}=-\frac{e^a+a e^a (x-1)+O\left((x-1)^2\right)}{(x-1)(x+1)}$$ $$y=-\frac{e^a}{2 (x-1)}+\frac{1}{4} (1-2 a) e^a+O\left((x-1)\right)$$ Now, consider the two cases $x\to 1^+$ and $x\to 1^-$.

Do the same around $x=-1$.