# 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