# Limits probably involving e^x-1/x special limit

## Homework Statement

lim of (a^x-a^-x-2)/(x^2) as x->0

Find the value of the limit^.

The answer is (lna)^2
I know how to get the answer using L'Hospital.
However, I do not want to use Hospital's rule.
I think I see the pattern with the special limit:
lim of (e^x-1)/x as x->0

But I just don't see how I can apply that. I tried to take ln() out of everything, but it leads me to nowhere.
Please help me.

Dick
Science Advisor
Homework Helper

## Homework Statement

lim of (a^x-a^-x-2)/(x^2) as x->0

Find the value of the limit^.

The answer is (lna)^2
I know how to get the answer using L'Hospital.
However, I do not want to use Hospital's rule.
I think I see the pattern with the special limit:
lim of (e^x-1)/x as x->0

But I just don't see how I can apply that. I tried to take ln() out of everything, but it leads me to nowhere.
Please help me.

a^x=e^(ln(a)x) is the connection between powers of e and a. You could also substitute the series expansions of the function. But I don't think knowing the special limit you've got there will help.

Then, how else could we solve this problem?
My school hasn't taught me Hospital's rule. So, there must be another way.

As far as I can see, L'Hopital's rule doesn't even apply since the numerator does not equal 0 if you plug in 0. a^0 - a^(-0) - 2 = 1 - 1 - 2 = -2.

You're simply dividing by a small positive number, so the limit is negative infinity.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
I think there's a typo in the original post. The problem should be
$$\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}.$$ You can rewrite this as
$$\lim_{x \to 0} \frac{a^{-x}[(a^x)^2 - 2a^x +1]}{x^2}.$$ You can do a little algebra, use what Dick said about expressing this in terms of ##e##, and change variables appropriately to relate it back to the special limit.

Dick
Science Advisor
Homework Helper
I think there's a typo in the original post. The problem should be
$$\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}.$$ You can rewrite this as
$$\lim_{x \to 0} \frac{a^{-x}[(a^x)^2 - 2a^x +1]}{x^2}.$$ You can do a little algebra, use what Dick said about expressing this in terms of ##e##, and change variables appropriately to relate it back to the special limit.

Ok, that's nice. Now I see how to use the 'special limit'.