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Thread: Doubt on a Precise Definition of Limit

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    Doubt on a Precise Definition of Limit

    Doubt on a Precise Definition of Limit-image010.jpg

    Please see the image. Why is 0 missing in the inequality as shown in the image. Any input is appreciated.

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    Nikunj
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    Re: Doubt on a Precise Definition of Limit

    Because $|f(x)-L|$ is positive (or zero) by definition. Zero is a permitted value as in the constant function, for example.

    In the first inequality, zero is not permitted because we are looked at where the function is headed, not the actual value of the function at the given point.
    Last edited by Archie; Jul 4th 2019 at 10:18 AM.
    Thanks from Nikunj and topsquark
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    Re: Doubt on a Precise Definition of Limit

    Thanks a ton! I didn't think about the constant function...wasted a lot of time today on this. Thanks again
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    Re: Doubt on a Precise Definition of Limit

    Quote Originally Posted by Nikunj View Post
    Click image for larger version. 

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ID:	39451Please see the image. Why is 0 missing in the inequality as shown in the image. Any input is appreciated.
    The statement that $\mathop {\lim }\limits_{x \to a} f(x) = L$ means that for $\varepsilon >0$ $\exists \delta>0$ so that if $0<|x-a|<\delta$ then $|f(x)--L|<\varepsilon~.$
    Have you ever asked yourself why $0<|x-a|<\delta~?$ That is to say WHY $\bf{\large 0<|x-a|~?}$
    Well, the limit as $x\to a$ means that the number $x$ is close to $a$ BUT NOT EQUAL TO $a$ or $0<|x-a|<\delta$
    That is the question you should have been asking!
    You should also note that in the question about the continuity of $f$ at $x=a$ we use $|x-a|<\delta$ . WHY?
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