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 $a$.
 A: It must be that
$$(1)\;\;\;ax-1+e^{bx}\xrightarrow [x\to 0]{}-1+1=0\;\;\text{(this is true for any values of }\;a,b)$$
$$(2)\;\;a+be^{bx}\xrightarrow[x\to 0]{}a+b=0\implies a=-b$$
Since applying l'Hospital we get the limit of the quotient of the derivatives...
Finally, and as you wrote, applying l'H a second time:
$$b^2e^{bx}\xrightarrow[x\to 0]{} b^2=2\implies b=\pm\sqrt 2\;,\;\;\text{so}\ldots$$
A: 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 $(a, b) = (-\sqrt{2}, \sqrt{2})$.
A: 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 $b^2 = 2$.
