y = x^(1/x)

has a pretty funky looking graph,
which rises sharply in the positive quadrant 1 then slowly decreases and then asymptotes to the x axis as x increases.

y = (1)^(1/1)= 1.0
y = (2)^(1/2)= 1.41421...
(square root of 2) y = (3)^(1/3)= 1.4422...
(cubed root of 3) y = (4)^(1/4)= 1.41421...
(the square root of the square root of 2 all squared, as it were.)
y = (5)^(1/5)= 1.3797...
y = (6)^(1/6)= 1.348...

and so on and so forth, to the nth root of n.

This function (y = x^(1/x); see the write-up above also) really is an interesting function. We can use L'Hopital's Rule to prove that this function approaches 1 as x goes to infinity. I use "lim" to stand for the limit as x goes to infinity.

Let y = lim  x     .


lim ln y = lim (- ln x)
                ln x  
         = lim  ----
         = lim  ---                  (by L'Hopital's Rule)
lim ln y = 0.

Thus lim y = e^0 = 1.

The limit of y as x approaches 0 from above is 0. This stems from the fact that the limit of ((1/x) * ln x) as x goes to 0 from above is negative infinity. (The limit of (1/x) is positive infinity and the limit of (ln x) is negative infinity; take the product).

With implicit differentiation, we can also find the maximum value of y, and the x-value at which y attains this maximum. (Note that we restrict x here to be positive.)

y = x

ln y = - ln x

 1 dy   1   1                1
 - -- = - * - + (ln x) * (- ---)             (right side by the chain rule)
 y dx   x   x               x^2

Factoring (-1/x^2) out of the right side, and multiplying both sides by y, we obtain

dy    y 
-- = --- * (1 - ln x).
dx   x^2

The derivative has only one root, at x = e (= 2.71828...), the sole x that makes (1 - ln x) equal to 0. (The other factor has no root.) The y value here is approximately 1.44467. Since this is the only relative extremum of the function and the limits to the left and right are below it, we can conclude that this is the maximum value of the function. (Looking at a graph of it may make this clearer.)

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