Thread: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

1. f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

This comes from Calculus with Analytic Geometry by Purcell and Varberg, fifth edition, section 2.8, chapter review true/false quiz #28.

If f(x) < 10 for all x, and limit as x-> 2 of f(x) exists then limit as x->2 of f(x) < 10.

I answered true because either the function is continuous and f(2) < 10 from the first condition or it is discontinuous and all function values of f(x) near x=2 are less than 10.

The answer key disagrees. What did I miss?

2. Re: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

Originally Posted by Zexuo
This comes from Calculus with Analytic Geometry by Purcell and Varberg, fifth edition, section 2.8, chapter review true/false quiz #28.

If f(x) < 10 for all x, and limit as x-> 2 of f(x) exists then limit as x->2 of f(x) < 10.

I answered true because either the function is continuous and f(2) < 10 from the first condition
True

or it is discontinuous and all function values of f(x) near x=2 are less than 10.
Also true, but that doesn't imply $lim_{x\to 2} < 10$ Consider$$f(x) = \begin{cases} -|x-10|,~x \ne 10\\ 0,~x=10 \end{cases}$$

3. Re: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

Aha! Thanks.

The following function would serve as a counter-example to the proposition in the exercise:
$$f(x) = \begin{cases} 10 - |5x - 10|, ~x \ne 2\\ 0, x = 2\end{cases}$$

4. Re: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

Yes. Although you got the idea I obviously should have been in bed sleeping at 11:00 pm instead of posting that example in #2.

5. Re: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

Originally Posted by Zexuo
This comes from Calculus with Analytic Geometry by Purcell and Varberg, fifth edition, section 2.8, chapter review true/false quiz #28.
If f(x) < 10 for all x, and limit as x-> 2 of f(x) exists then limit as x->2 of f(x) < 10.
I answered true because either the function is continuous and f(2) < 10 from the first condition or it is discontinuous and all function values of f(x) near x=2 are less than 10.
The answer key disagrees. What did I miss?
Rearranging my home office, I found the sixth edition of Purcell and Varberg. So I looked that problem.
There it was and my note was different from their counter-example.
$f(x)=\begin{cases}10-|x-2| & x\ne 2 \\ 2 & n= 2\end{cases}$

The reason I like this particular form is that the use of $|x-2|$ works so well with the definition of continuity.
If $x\ne 2$ then $|x-2|>0$, so that $10-|x-2|<10$ But clearly $\displaystyle \mathop {\lim }\limits_{x \to 2} \left[ {10 - |x - 2|} \right] = 10$.
We can argue that $\displaystyle \mathop {\lim }\limits_{\delta \to 0} \left[ {10 - \delta } \right] = 10$