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

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

    Quote Originally Posted by Zexuo View Post
    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$$
    f(x) = \begin{cases}
    -|x-10|,~x \ne 10\\
    0,~x=10 \end{cases}$$
    Thanks from Archie, Zexuo and topsquark
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    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:
    $$f(x) = \begin{cases}
    10 - |5x - 10|, ~x \ne 2\\
    0, x = 2\end{cases}$$
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    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.
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    Re: f(x) < 10 {x:x}, limit as x->2 exists, then limit as x->2 < 10?

    Quote Originally Posted by Zexuo View Post
    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 $\displaystyle \mathop {\lim }\limits_{\delta \to 0} \left[ {10 - \delta } \right] = 10$
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