How do you prove that #sqrtx# is continuous?
1 Answer
We need to prove that for any point
# | x-a| < delta => | sqrt(x)-sqrt(a) | < epsilon #
So, to find a suitable
# \ \ \ \ \ | 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
# \ \ \ \ \ \ a−1 < x < a+1#
# :. sqrt(x) < sqrt(a+1)# .
# :. sqrt(x) + sqrt(a) < sqrt(a+1) + sqrt(a)# ,
which combined with [1] gives;
So, let
Hence we have proved that