<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] As the limit in question is of the indeterminate form "$\frac{0}{0}$", we use L'Hôpital's rule and consider the limit $$\lim_{x \to 0} \frac{a +be^{bx}}{2x}.$$ Note that $\lim_{x\to 0}a+be^{bx} = a+be^0 = a+b$ and $\lim_{x\to 0} 2x = 0$. The only way the overall limit can be $1$ is if we have a limit in the indeterminate form "$\frac{0}{0}$". For this to occur, we need $a = - b$. Then, using L'Hôpital's rule once more, we consider the limit $$\lim_{x\to 0}\frac{b^2e^{bx}}{2} = \frac{b^2}{2}.$$ As we want this limit to be $1$, we find that $b = \pm\sqrt{2}$ and hence $a = \mp\sqrt{2}$.
That is, $$\lim_{x\to 0}\frac{a-1+e^{bx}}{x^2} = 1$$ when $(a, b) = (\sqrt{2}, -\sqrt{2})$ or $(a, b) = (-\sqrt{2}, \sqrt{2})$.
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