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Math Help - The Basics of Limits -- Calculus Intro

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    The Basics of Limits -- Calculus Intro

    So the question is (and sorry for butchering this... I have tried Latex but cannot figure it out for the life of me)

    limit[x:1-,-3x+2] (the limit on (-3x+2) as x approaches 1 from the left)
    (PS. Tell me if this notation is acceptable, or my wording is bad...)

    The book we use provides solutions, and the solutions show simply subbing in 1 to the (-3x+2), but that wouldn't work if it was piecewise, right?

    So how would you do this other than using values like .999999 ?
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    you're finding a one-sided limit, so no problem ... do like the book says.

    \lim_{x \to 1^-} (-3x+2) = -1
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    Okay, I understadn that as long as the domain is not restricted, that works out fine... But what if it was restricted so that x>1 ..... Then this wouldn't work out, correct? (And to tell you the WHOLE truth, that is what the book does... The question is actually a piecewise, and it just subs in 1 from both sides)
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    seems like some information is missing here ...

    how is the piece-wise function defined in the first place, and what limit does the problem want you to find?

    working with specifics is easier to explain.
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    I am sorry, I do not have the book on me anymore. I will likely check with the instructor tomorrow but... I'll see if I remember this all.

    essentially it was:

    f(x)= {-3x+2}, x<1 and {x^2}, x>=1

    The questions asks us to find the limits from both directions, at x=1 ...

    Then the SOLUTION shows them subbing 1 into the first side of the piecewise function .... That doesn't seem right to me.... 1 is not in the domain of that question so subbing it into there will not return the correct y value.
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    Quote Originally Posted by mike_302 View Post
    I am sorry, I do not have the book on me anymore. I will likely check with the instructor tomorrow but... I'll see if I remember this all.

    essentially it was:

    f(x)= {-3x+2}, x<1 and {x^2}, x>=1

    The questions asks us to find the limits from both directions, at x=1 ...

    Then the SOLUTION shows them subbing 1 into the first side of the piecewise function .... That doesn't seem right to me.... 1 is not in the domain of that question so subbing it into there will not return the correct y value.
    ok, so here's the thing. with limits, we don't actually care about being "at" the point, but about being "close" to the point. now \lim_{x \to 1^-} f(x) is asking, when x gets "close" to 1 coming from the left (meaning, the x-values are always a little less than 1 but can get arbitrarily close to 1) what value does f(x) get "close" to?

    now, your piecewise function is defined as -3x + 2 for x < 1. so your left hand limit is concerned with this definition of f(x). now, as long as the funciton is continuous, you can just plug in the value. so you plug in 1 into -3x + 2


    Now, if they asked for \lim_{x \to 1^+}f(x), they are asking for the limit as x gets close to one from the right. hence you look at values where x is a little bigger than 1. so you look for how f(x) is defined for x > 1. here, we have f(x) = x^2 for x \ge 1, so that

    \lim_{x \to 1^+}f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1
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    also note ... if the left limit = the right limit, then the overall limit exists and has the value equal to the left and right limits.
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