Show that ##1 - \sqrt {1 - x^2}## is continuous on the interval ##[-1, 1]##

[Callumnc1]

Homework Statement:

Please see below

Relevant Equations:

Please see below

For this problem,

I don't understand why they are saying $-1 < a < 1$ since they are trying to find where $f(x)$ is continuous including the endpoints $f(-1)$ and $f(1)$

Why is it not: $-1 ≤ a ≤1$

Many thanks!

 

[PeroK]

Homework Statement:: Please see below
Relevant Equations:: Please see below

For this problem,
View attachment 322317
I don't understand why they are saying $-1 < a < 1$ since they are trying to find where $f(x)$ is continuous including the endpoints $f(-1)$ and $f(1)$

Why is it not: $-1 ≤ a ≤1$

Many thanks!

Because they forgot about the endpoints.

 

[Callumnc1]

Because they forgot about the endpoints.

Thank you for your reply @PeroK!

 

[PeroK]

Thank you for your reply @PeroK!

They should have done one-sided limits at the end points, in addition to two sided limits at the interior points. As any good maths student will tell you!

 

[FactChecker]

The part of the proof that you show matches the first line: "If $-1 \lt x \lt 1$". Is there another part of the proof that you have not shown? If not, then they just made a mistake and left it out.

 

[Callumnc1]

Thank for your replies @PeroK and @FactChecker !

There was a couple of lines that I did not screen shot,

However, did they really need those lines if they had just said $-1 ≤ a ≤1$?

Many thanks!

 

[FactChecker]

However, did they really need those lines if they had just said $-1 ≤ a ≤1$?

The theorems and laws that are referenced in the first part are probably stated only for two-sided limits, so it would not be valid to include the endpoints that are one-sided limits. By separating the endpoints, they can use the phrase "similar calculations" and infer similar one-sided theorems and laws.

 

[PeroK]

Thank for your replies @PeroK and @FactChecker !

There was a couple of lines that I did not screen shot,
View attachment 322319

However, did they really need those lines if they had just said $-1 ≤ a ≤1$?

Many thanks!

So, they didn't forget about the endpoints after all!

 

[Callumnc1]

The theorems and laws that are referenced in the first part are probably stated only for two-sided limits, so it would not be valid to include the endpoints that are one-sided limits. By separating the endpoints, they can use the phrase "similar calculations" and infer similar one-sided theorems and laws.

Thank you for your replies @FactChecker and @PeroK!

I think I'm starting to understand. So basically, you can't take the limits of the end points, so you just take the right- and left-hand limits to prove it is continuous.

However, I though you could not do that since the text also states that in order for a function to be continuous at a number a:

However, for the end points they only took the right hand or left hand limit for reach end point. How dose that me it is continuous at $x = -1, 1$ (since the limits at each of those end points DNE)?

For example, for $x = 1$ You cannot take the right-hand limit since there is no graph there (so left-hand limit dose not equal right-hand limit, so limit DNE).

I think this could be something to do with Definition 3.

Many thanks!