$\displaystyle Find\ all\ (a,b)\ such\ that$

$\displaystyle \lim_{x \rightarrow 0} \frac{ax-1+e^{bx}}{x^2}=1$

I can find b by using L'Hopital's Rule, where b is $\displaystyle \pm \sqrt{2}$, but I don't know how to find a

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- May 2nd 2010, 12:45 AMacevipaLimits question
$\displaystyle Find\ all\ (a,b)\ such\ that$

$\displaystyle \lim_{x \rightarrow 0} \frac{ax-1+e^{bx}}{x^2}=1$

I can find b by using L'Hopital's Rule, where b is $\displaystyle \pm \sqrt{2}$, but I don't know how to find a - May 2nd 2010, 01:15 AMDebsta
- May 2nd 2010, 01:52 AMacevipa
- May 2nd 2010, 03:09 AMHallsofIvy
I'm not clear how you can use L'Hopital and NOT get a. Differentiating both numerator and denominator, gives $\displaystyle \frac{a+ be^{bx}}{2x}$ and using L'Hopital's rule again, $\displaystyle \frac{b^2e^{bx}}{2}$ which has a limit of $\displaystyle \frac{b^2}{2}$. In order for that to be 1, b must be $\displaystyle \pm\sqrt{2}$.

**But**, in order that we be**able**to use L'Hopital a second time, on the fraction $\displaystyle \frac{a+ be^{bx}}{2x}$, since the denominator goes to 0, the numerator must also- we must have $\displaystyle \lim_{x\to 0}a+ be^{bx}= a+ b= 0$. That is a= -b. - May 2nd 2010, 03:31 AMDebsta