Find : limit of as x->0+ x^x?

When a question asks for the limit as X -> 0^+ or as X -> 0^-, it is referring to the direction of approach of the limit.

First off, understand the general idea of what a limit is. A limit of a function as it approaches a value of x is NOT the same as the value of the function at that value x. For example, consider this function:

f(x) = (x^2)/x

By looking at the graph of f(x), or by comparing the results of f(x) for values of x close to 0, it is clear that the function is getting closer and closer to 0 as x gets closer to 0. However, if you plug in 0, you would get an undefined result. Thus, there is a hole at x = 0 on the graph of f(x).

When you see the ^+ or ^- in a limit, it is asking whether you are approaching from the right (from positive infinity) or left (from negative infinity) side, respectively. This can be visualized on a graph, as the right side of a function is closer to positive infinity, and the left side is closer to negative infinity.

Consider this function:

g(x) = 1/x

When you graph this function, or look at the results when you substitute x values that are very close to zero, you will notice that very small positive x-values (such as .00001) bring g(x) closer to positive infinity, while very small negative x-values (such as -.000001) bring g(x) closer to negative infinity. Thus, to say that the limit as x -> 0 of g(x) is infinity is incorrect, because on the left side of x = 0 it approaches negative infinity, and to say the limit is negative infinity is incorrect for the opposite reason.

Therefore, when you specify the + or - in the infinity, you define which side you are approaching from. So, for example, the limit of g(x) as x -> 0^+ is positive infinity, and as x ->^- is negative infinity.

To answer the actual question; if you look at the graph of x^x, you will notice that it is continuous - it is not interrupted or cut off at any point. Thus, the limit of the function as x -> 0 from both sides is equal. Also, x^x is defined at x = 0. So you can substitute in x = 0, making it 0^0. Since exponential rules state that any number raised to 0 is equal to 1, the answer is 1.

You have to take the logarithmic limit.

ln (lim x->0+ (x^x) ) = lim x->0 (ln x^x) = lim x->0 (x ln x)

Using l'Hopital's, lim x ln x = lim (ln x) / (1/x) = lim (1/x) / (-1/x^2) = 0

So ln (the limit) = 0, and (the limit) = e^0 = 1