If p is a polynomial, then lim x ->b p(x)=p(b).... Why is this true?

[Integrate]

My understanding is that limits do not equal the value of a function. They only show the behavior AROUND the value of a function. It was made very clear to me that limits never equal the output value.

Why is this not the case here?

James Stewart 7th edition

 

[pka]

It was made very clear to me that limits never equal the output value.
James Stewart 7th edition

Whoever made that clear to you did you an enormous disservice. It is not the case.
If a function [imath]f[/imath] is continuous at [imath]x_0[/imath] then [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) = \left( {{x_0}} \right)[/imath]
Now what you may have been told that in evaluating limits do not simply substitute.
i.e. [imath]\mathop {\lim }\limits_{x \to {a}} f(x)[/imath] may well exist but it may be that [imath]\mathop {\lim }\limits_{x \to {a}} f(x)\ne f(a)[/imath]

[imath][/imath][imath][/imath][imath][/imath]

 

[BigBeachBanana]

If a function [imath]f[/imath] is continuous at [imath]x_0[/imath] then [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) =\red{ \left( {{x_0}} \right)}[/imath]

I think you meant [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) = f\left({{x_0}}\right)[/imath]

 

[Dr.Peterson]

View attachment 32416



My understanding is that limits do not equal the value of a function. They only show the behavior AROUND the value of a function. It was made very clear to me that limits never equal the output value.

Why is this not the case here?

James Stewart 7th edition

Check in your book for the definition of continuity. There should also be a theorem that says a polynomial is continuous everywhere.

Together, these answer the question and correct your misunderstanding.

As to the definition of limits, here is a snippet from what may be your book:

​

This is probably what you misinterpreted. Do you see the difference?

 

[Integrate]

Whoever made that clear to you did you an enormous disservice. It is not the case.
If a function [imath]f[/imath] is continuous at [imath]x_0[/imath] then [imath]\mathop {\lim }\limits_{x \to {x_0}} f(x) = \left( {{x_0}} \right)[/imath]
Now what you may have been told that in evaluating limits do not simply substitute.
i.e. [imath]\mathop {\lim }\limits_{x \to {a}} f(x)[/imath] may well exist but it may be that [imath]\mathop {\lim }\limits_{x \to {a}} f(x)\ne f(a)[/imath]

[imath][/imath][imath][/imath][imath][/imath]

Why can they not substitute?

 

[Integrate]

Check in your book for the definition of continuity. There should also be a theorem that says a polynomial is continuous everywhere.

Together, these answer the question and correct your misunderstanding.

As to the definition of limits, here is a snippet from what may be your book:

View attachment 32419​

This is probably what you misinterpreted. Do you see the difference?

Thank you and yes this is my book.

Let me see if I understand this then.

If a limit and a function equal each other and therefore it is continuous.

BUT for a limit to exist it does not need to actually equal f(a). Just that both sides approach f(a).

 

[Dr.Peterson]

Thank you and yes this is my book.

Let me see if I understand this then.

If a limit and a function equal each other [then] it is continuous.

BUT for a limit to exist it does not need to actually equal f(a). Just that both sides approach f(a).

Correct. But if you know the function is continuous, then you can just use f(a).

Now what you may have been told that in evaluating limits do not simply substitute.
i.e. [imath]\mathop {\lim }\limits_{x \to {a}} f(x)[/imath] may well exist but it may be that [imath]\mathop {\lim }\limits_{x \to {a}} f(x)\ne f(a)[/imath]

Why can they not substitute?

You can substitute, under the right conditions, namely when you know the function is continuous.

What he said you can't do is "simply" substitute -- that is, do so blindly without checking whether it is valid.

The error in your statement "My understanding is that limits do not equal the value of a function" was the omission of a single word: "My understanding is that limits do not necessarily equal the value of a function."

 

[pka]

Why can they not substitute?

Lets take a really simple example.
[imath]f(x)=|x-1|+3\text{ if }x\ne 1~\&~4\text{ if }x=1 [/imath]
In this example [imath]\mathop {\lim }\limits_{x \to 1} f(x) = 3[/imath] BUT [imath]\bf f(1)=4[/imath]

[imath][/imath]

 

[Integrate]

Lets take a really simple example.
[imath]f(x)=|x-1|+3\text{ if }x\ne 1~\&~4\text{ if }x=1 [/imath]
In this example [imath]\mathop {\lim }\limits_{x \to 1} f(x) = 3[/imath] BUT [imath]\bf f(1)=4[/imath]

[imath][/imath]

Excellent example. Thank you.

 

[Integrate]

Correct. But if you know the function is continuous, then you can just use f(a).


You can substitute, under the right conditions, namely when you know the function is continuous.

What he said you can't do is "simply" substitute -- that is, do so blindly without checking whether it is valid.

The error in your statement "My understanding is that limits do not equal the value of a function" was the omission of a single word: "My understanding is that limits do not necessarily equal the value of a function."

This clears up all my confusion. Thank you greatly.