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Math Help - Limits/continuity

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    Limits/continuity

    Hi,

    First post, was not sure where to start with this one. I'm looking for an example of a function that is defined for all real numbers and for which the limit as x approaches 0 exists, but the limit as x approaches 0 is not the same as when the function f(0).

    IIRC this would mean it is discontinuous (removable I think?) so would any function that when direct substitution is used returns 0/0 (and requires factoring, common denominator, conjugate, etc to find its limit) work for this? or would that still not be defined for all real numbers I guess, I'm kind of lost on where to start.
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    Re: Limits/continuity

    What about \displaystyle \begin{align*} f(x) =  \begin{cases} \frac{x^2 + 3x + 2}{x + 1} \textrm{ if } x \neq -1 \\ 5 \textrm{ if } x = -1  \end{cases} \end{align*}
    Thanks from Holly
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    Re: Limits/continuity

    Thanks for the response, this is where I get confused though - the first one would be 2/0 and dividing by 0 makes it undefined, so would that disqualify the "function that is defined for all numbers" part? Also x would be approaching 0, not -1, unless i'm looking at something differently here?
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    Re: Limits/continuity

    Quote Originally Posted by Holly View Post
    Thanks for the response, this is where I get confused though - the first one would be 2/0 and dividing by 0 makes it undefined, so would that disqualify the "function that is defined for all numbers" part? Also x would be approaching 0, not -1, unless i'm looking at something differently here?
    It's a hybrid function, we have realised that the first part of the function is undefined at the point x = -1 so we have defined the function to have a value at that point.

    And of course, you need to evaluate the limit as x goes to -1, because that's the point of discontinuity.
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    Re: Limits/continuity

    Is -1 coming from anything specific though or is it just a random number? I know it makes the first 0, but using 0 itself does too, so is there any other reason why -1 is used?

    Sorry, just having trouble wrapping my head around these concepts in general.
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    Re: Limits/continuity

    Quote Originally Posted by Holly View Post
    Is -1 coming from anything specific though or is it just a random number? I know it makes the first 0, but using 0 itself does too, so is there any other reason why -1 is used?

    Sorry, just having trouble wrapping my head around these concepts in general.
    No, x = -1 does not make the first function 0, it makes the DENOMINATOR of the first function 0, which gives the point of discontinuity.

    x = 0 does NOT make the denominator 0, so is NOT a point of discontinuity.
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    Re: Limits/continuity

    The simplest such example is just f(x)= 1 if x\ne 0, f(0)= 0.
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