# Limits question

• May 2nd 2010, 12:45 AM
acevipa
Limits question
$Find\ all\ (a,b)\ such\ that$

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

I can find b by using L'Hopital's Rule, where b is $\pm \sqrt{2}$, but I don't know how to find a
• May 2nd 2010, 01:15 AM
Debsta
Quote:

Originally Posted by acevipa
$Find\ all\ (a,b)\ such\ that$

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

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

There is no restriction on what a can be. a is an element of R. This is clear when you use L'Hopital twice.
• May 2nd 2010, 01:52 AM
acevipa
Quote:

Originally Posted by Debsta
There is no restriction on what a can be. a is an element of R. This is clear when you use L'Hopital twice.

$(a,b)=(-\sqrt{2}, \sqrt{2})\ or\ (\sqrt{2}, -\sqrt{2})$
• May 2nd 2010, 03:09 AM
HallsofIvy
Quote:

Originally Posted by acevipa
$Find\ all\ (a,b)\ such\ that$

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

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

I'm not clear how you can use L'Hopital and NOT get a. Differentiating both numerator and denominator, gives $\frac{a+ be^{bx}}{2x}$ and using L'Hopital's rule again, $\frac{b^2e^{bx}}{2}$ which has a limit of $\frac{b^2}{2}$. In order for that to be 1, b must be $\pm\sqrt{2}$.

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

Originally Posted by HallsofIvy
I'm not clear how you can use L'Hopital and NOT get a. Differentiating both numerator and denominator, gives $\frac{a+ be^{bx}}{2x}$ and using L'Hopital's rule again, $\frac{b^2e^{bx}}{2}$ which has a limit of $\frac{b^2}{2}$. In order for that to be 1, b must be $\pm\sqrt{2}$.

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

Yes my mistake. I overlooked the fact that we must have 0/0 for L'Hopital's rule to apply. Sorry for the confusion.