Question

How do you prove that `sqrtx` is continuous?

Answer 1

We need to prove that for any point `a in (0,oo)`, for every `epsilon gt 0` there exists a `delta gt 0` such that

`| x-a| < delta => | sqrt(x)-sqrt(a) | < epsilon`

So, to find a suitable `delta`, we must look at the inequality `|sqrt(x)-sqrt(a)| < epsilon`. Since we want an expression involving `|x−a|`, then multiply by the conjugate to remove the square roots, so:

`\ \ \ \ \ | sqrt(x)-sqrt(a) | < epsilon`
`:. | sqrt(x)-sqrt(a) | * | sqrt(x)+sqrt(a) | < epsilon * | sqrt(x)-sqrt(a) |`
`:. | (sqrt(x)-sqrt(a)) * (sqrt(x)+sqrt(a)) | < epsilon * | sqrt(x)-sqrt(a) |`
`:. | sqrt(x)sqrt(x) +sqrt(x)sqrt(a)-sqrt(x)sqrt(a)-sqrt(a)sqrt(a)| < epsilon * | sqrt(x)-sqrt(a) |`
`:. | x -a| < epsilon * | sqrt(x)-sqrt(a) |` ..... [1]

Now, if you require that `|x−a|<1`, then it follows that `x−a < 1`, so:

`\ \ \ \ \ \ a−1 < x < a+1`
`:. sqrt(x) < sqrt(a+1)`.
`:. sqrt(x) + sqrt(a) < sqrt(a+1) + sqrt(a)`,

which combined with [1] gives;

`|x−a| < epsilon (sqrt(a+1) + sqrt(a))`

So, let `delta = min(1, epsilon (sqrt(a+1) + sqrt(a)))`.

Hence we have proved that `f(x)=sqrt(x)` is continuous on `(0,oo)`