<s>[INST] Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$ Find $(a, b)$ such that $\lim_{x \to 0} \frac{ax -1 + e^{bx}}{x^2} = 1$
I am able to find $b = \pm \sqrt{2}$ using L'Hopitals Rule, but unable to do anything for $a$.
 [/INST] Another way to do it is with Taylor expansion of the numerator,
$$-1 + ax + (1 + bx + \frac{b^2}{2}x^2 + \mbox{higher order terms}) = (a+b)x + \frac{b^2}{2} + \mbox{ h.o.t.}).$$
Now dividing by $x^2$ and sending $x \to 0$ it is clear that the higher order terms vanish, and $a = -b$ to make sure the limit exists. We are left with 
$$\lim_{x\to 0} \frac{b^2}{2} \frac{x^2}{x^2} = \frac{b^2}{2}.$$
And as such $b^2 = 2$.
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