How do you prove that #sqrtx# is continuous?

1 Answer
Steve M
Dec 14, 2016

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)#

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