Prove, without using L'Hopitals rule, that:

lim x® 0

((2x+1)/2)1/x=21/2

Thank you!

Pooya,

Here's a way, if we assume the function to be continuous at 0 (i.e. it does indeed converge to some limit at all).

Let f(x) = [(2^x +1)/2]^1/x

Now f(-x) = [(2^-x +1)/2]^-1/x

So f(x)f(-x) = [(2^x +1)/(2^-x +1)]^1/x = 2

So the limit as x-> 0 of f(x)f(-x) is 2.

But as the function is continuous at 0, the limit of f(x)f(-x) is equal to the limit of f(x)² . So f(x)² tends to 2, and f(x) tends to sqrt(2).

Perhaps someone can finish what I've started.

Yours,

Michael

Nice proof Michael :)

I liked it Michael. Very nice!