Question

How do you prove that the function `f(x) = x^(1/2)` is continuous at x=1/2?

Answer 1

There are many ways to prove this - one could use :

1. Epsilon-delta methods
   `y=f(x)` is continuous at `x=x_0iff AA epsilon >0, EE delta>0` such that `|x-x_0| < delta=> |f(x)-f(x_0)| < epsilon`

2. Topological methods
   `y:X->Y;y=f(x)` is continuous everywhere (specifically at `x_0 in X` iff and only if for every open subset A in Y, the inverse function `f^(-1)(A)` is open in X.

3. Functional analysis methods using sequential criterion .
   `y=f(x)` is continuous at `x=x_0iff` for every sequence `(x_n)` converging to `x_0`, the sequence of function values `(f(x_n))` converges to `f(x_0)`.

I shall prove the continuity using the latter definition in this case :

Let `(x_n)=(1/2+1/n^2)` be a sequence in `RR`.
Since `1/n^2->0=>(x_n)->1/2 in RR`.

But `(f(x_n))=(1/2+1/n^2)^(1/2)` which converges to `(1/2)^(1/2)= f(1/2)`

`therefore f` is continuous at `x=1/2`.