Finding the Ordered Pairs for an Exponential Limit

[erisedk]

Homework Statement

If $a, b \in \{1,2,3,4,5,6\}$, then number of ordered pairs of $(a,b)$ such that $\lim_{x\to0}{\left(\dfrac{a^x + b^x}{2}\right)}^{\frac{2}{x}} = 6$ is

Homework Equations

The Attempt at a Solution

So, this is a typical exponential limit.

$\lim_{x\to0}e^{\frac{2}{x}.\ln\left(\frac{a^x + b^x}{2}\right)} = 6$

Using L'Hospital

$\lim_{x\to0}e^{2.\frac{2}{a^x + b^x}.(a^x\ln a + b^x\ln b)} = 6$

This on substituting the limit simplifies to

$e^{2.(\ln a + \ln b)} = 6$

$e^{\ln {(ab)}^2} = 6 \Rightarrow {(ab)}^2 = 6$ However, the answer only has $ab = 6$. What's wrong?

 

[PeroK]

You can't use L'Hopital's rule there!

That said, I see you're using it inside a continuous function, which I guess is a generalisation of L'Hopital. You just made a simple error with your differentiation.

 

[erisedk]

Oh, yesss, I see it now. That was rather dumb. Thank you!