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?
 A: I think, your limit is equal to $0$ for all reals $a$ and $b$ because
$$\lim_{x\rightarrow+\infty}\frac{e^{-2x}}{x^2}=0$$ and
$$\lim_{x\rightarrow+\infty}\frac{1+ax}{x^2(1+bx)}=0.$$
I think, much more interesting question with $x\rightarrow0.$
If so, since$$\left(e^{-2x}-\frac{1+ax}{1+bx}\right)_{x=0}=0,$$  we need 
$$\left(e^{-2x}-\frac{1+ax}{1+bx}\right)'_{x=0}=0$$ or
$$\left(-2e^{-2x}-\frac{a-b}{(1+bx)^2}\right)_{x=0}=0,$$ which gives
$$a-b=-2.$$
Also, we need 
$$\left(-2e^{-2x}-\frac{a-b}{(1+bx)^2}\right)'_{x=0}=0$$ or
$$\left(4e^{-2x}+\frac{2b(a-b)}{(1+bx)^3}\right)_{x=0}=0,$$
which gives $b=1$ and $a=-1.$
A: 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


*

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

*$b-a-2=0 \implies a=-1$
A: The question might have been the limit as $x\to0$ instead.
Then you can apply L'Hopital's Rule to 
$$\frac{(1+bx)e^{-2x}-(1+ax)}{(1+bx)x^2}$$
A: The given limit condition is equivalent to $$\lim_{x\to 0}\frac{(1+bx)-(1+ax)e^{2x}}{x^2}=0\tag{1}$$ We need to make use of the following limits $$\lim_{x\to 0}\frac{e^x-1}{x}=1,\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}\tag{2}$$ The first one is standard and the second one is derived from first one via L'Hospital's Rule.
By splitting the numerator in $(1)$ appropriately we can rewrite $(1)$ as $$\lim_{x\to 0}\frac{1+2x-e^{2x}}{(2x)^2}\cdot 4+\frac{b-a-2} {x} -a\cdot\frac{e^{2x}-1} {2x}\cdot 2=0$$ And using limits in $(2)$ we see that the above is equivalent to $$\lim_{x\to 0}\frac{b-a-2}{x}=2+2a$$ Thus we have $b-a=2$ and $2+2a=0$ so that $a=-1,b=1$.
