# Limit problem

## 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

## 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?

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PeroK
Homework Helper
Gold Member
2020 Award
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.

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