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?

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 $a=-1,b=1$.
