The limit of the expression (a^x - a^-x - 2)/(x^2) as x approaches 0 evaluates to (ln(a))^2. While L'Hospital's rule is a common method for solving this limit, alternative approaches involve recognizing the connection between exponential functions and their logarithmic counterparts. Specifically, substituting a^x with e^(ln(a)x) and applying series expansions can lead to the correct evaluation. A critical correction to the original problem statement is necessary, changing the numerator to (a^x + a^-x - 2) for accurate analysis.
PREREQUISITESStudents studying calculus, particularly those focusing on limits and exponential functions, as well as educators seeking alternative teaching methods for limit evaluation.
alingy1 said: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.
vela said: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.