how to find a, b that satisfy $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ How can I find those $a$ and $b$ in  $\displaystyle \lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ ?
[my attempt] Since the denominator is $x^2$, it will be $0$. And $e^{-2x}$ is 1 so I get $1-\frac{1+ax}{1+bx} = 0$. I get $a=b$ but Im not sure this is right. Am I taking a wrong way?

As already notice the case $\lim_{x \to \infty} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$ is trivial for $$\lim_{x \to 0} \frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}=0$$
we have by $e^{-2x}=1-2x+2x^2+o(x^2)$
$$\frac{e^{-2x} -\frac{1+ax}{1+bx}}{x^2}
=\frac{e^{-2x}(1+bx) -(1+ax)}{x^2(1+bx)}
=\frac{(1-2x+2x^2+o(x^2))(1+bx) -(1+ax)}{x^2+o(x^2)}=$$
$$=\frac{(b-a-2)x+(2-2b)x^2+o(x^2)}{x^2+o(x^2)}$$
and the limit is zero for


*

*$2-2b=0 \implies b=1$

*$b-a-2=0 \implies a=-1$
