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Math Help - Limits and function values

  1. #1
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    Limits and function values

    Suppose we wish to compute

    \lim_{x \rightarrow 1}  \frac{x^2 + x + 1}{x + 3} \, ,

    where \small f(x) = \frac{x^2 + x + 1}{x + 3}. Since limits only care about what happens as we approach x = 1, why do we then compute the limit by simply plugging in x = 1 in f(x)?

    What if we were to discover that f(1) = 1337 but f(0.99999) going towards negative infinity and f(1.0000001) going towards \pi, or something random like that? How can we be so sure of this not happening (without actually checking values close to x = 1)?

    What's the justification for equating \lim_{x\rightarrow 1} with f(1)?
    Last edited by MathCrusader; January 16th 2013 at 02:40 PM.
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  2. #2
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    Re: Limits and function values

    Quote Originally Posted by MathCrusader View Post
    What's the justification for equating \lim_{x\rightarrow 1} with f(1)?
    It is that f(x) is continuous at x= 1; therefore, its limit equals its value. The fact that f(x) is continuous is implied by the fact that f(x) is a composition of continuous functions. In particular, it is known that 1 / (x + 3) is continuous everywhere except when the denominator turns into 0.
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  3. #3
    Junior Member Barioth's Avatar
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    Re: Limits and function values

    If i remember right, if the fonction is continuous in f(a) then (lim x=> a) = f(a)
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  4. #4
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    Re: Limits and function values

    Quote Originally Posted by emakarov View Post
    It is that f(x) is continuous at x= 1; therefore, its limit equals its value. The fact that f(x) is continuous is implied by the fact that f(x) is a composition of continuous functions. In particular, it is known that 1 / (x + 3) is continuous everywhere except when the denominator turns into 0.
    Ah, alright I understand! But suppose we didn't know that f(x) was continuous?
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  5. #5
    Junior Member Barioth's Avatar
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    Re: Limits and function values

    well you cant not know... you have to check it to evaluate the limit. You probably do it without even knowing.
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