Expert: Paul Klarreich - 12/3/2006

Question
Prove that
lim as x->a (sqrt x)= sqrt a
I just cant figure out the right epsilon delta combination.

Also can you prove that lim x->a of x^n=a^n
I think this one involves the bernoulli inequality.
Thankyou.

Answer
Questioner:  sammy
Category:  Calculus
 
Question:  Prove that
lim as x->a (sqrt x)= sqrt a
I just cant figure out the right epsilon delta combination.

Also can you prove that lim x->a of x^n=a^n
I think this one involves the bernoulli inequality.
Thank you.
...............................
Hi, Sammy,

It has been a while since I did things like this, but maybe the following will work out:
[I will write  'e' and 'd' in this reply -- can't make epsilons and deltas.]

You want to prove this limit:
 
lim  sqrt(x) = sqrt(a)
x->a

OR, given e > 0,  find  d > 0 such that whenever |x - a| < d,

|sqrt(x) - sqrt(a)| < e.

Now if we note that a bit of rationalizing might help:  Multiply both sides by (sqrt(x)

+ sqrt(a)):

| x - a | < e (sqrt(x) + sqrt(a))

That might give us the clue.  

| x - a | = | sqrt(x) - sqrt(a) | (sqrt(x) + sqrt(a))

Now when x is near x, sqrt(x) is near sqrt(a), and the second factor is near 2 sqrt(a).  

If it is near  2 sqrt(a), we can require that it be greater than 1 sqrt(a) or sqrt(a)

So if we simply choose  d = sqrt(a) e, we will have:

| x - a | < d

| x - a | < e sqrt(a)

| x - a | < e (sqrt(x) + sqrt(a))

   | x - a |         
--------------------  < e
(sqrt(x) + sqrt(a))


|sqrt(x) - sqrt(a)| (sqrt(x) + sqrt(a))
---------------------------------------  < e
        (sqrt(x) + sqrt(a))


|sqrt(x) - sqrt(a)| < e

Which is what we wanted to prove.
...........................................
About:

lim  x^n = a^n
x->a

Also can you prove that lim x->a of x^n=a^n
I think this one involves the bernoulli inequality.

Try this approach:

x^n - a^n  has  (x-a) as a factor. [Remember your Factor Theorem and Remainder Theorem

from precalculus?]

x^n - a^n = (x - a)(x^n-1 + x^n-2 a + x^n-3 a^2 + ... + a^n-1)

[I'm being a bit sloppy with the parenthesizing.]

Now again, suppose we make a restriction on x.  It's supposed to be near a, right?  Then

assume it is less than 2a.  Then we have:

x^n - a^n < (x - a)(2^n-1 a^n-1 + 2^n-2 a^n-2 a + 2^n-3 a^n-3 a^2 + ... + a^n-1)

x^n - a^n < (x - a)(2^n-1 a^n-1 + 2^n-2 a^n-1   + 2^n-3 a^n-1     + ... + a^n-1)

x^n - a^n < (x - a)(2^n-1 + 2^n-2  + 2^n-3  + ... + 1)a^n-1

x^n - a^n < (x - a)(n 2^n-1)a^n-1

Now let  d = e/(n 2^n-1)a^n-1, and then

|x - a| < e/(n 2^n-1)a^n-1 and so:

x^n - a^n < (n 2^n-1)a^n-1[e/(n 2^n-1)a^n-1] = e