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anonymous

  • 4 years ago

can i get some help with limits?

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  1. KingGeorge
    • 4 years ago
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    Almost certainly. What kind of limits do you need help with?

  2. anonymous
    • 4 years ago
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    i have to show that f(x) has a removable discontiunty at x=2 and for what value of f(2) f(x) countionous. the equation is \[(2x^2+3x-14)/(x-2)\]

  3. anonymous
    • 4 years ago
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    check to see if 2 makes the numerator zero. if it does factor and cancel

  4. anonymous
    • 4 years ago
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    you get \[(x-2)(2x+7)\] canceling gives you \[2x+7\] replace x by 2 and get 11

  5. anonymous
    • 4 years ago
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    i don't really get continuity at all i guess

  6. anonymous
    • 4 years ago
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    ok the algebra is easy enough yes? so now the question is, what is the difference between \[f(x)=\frac{2x^2+3x-14}{x-2}\] and \[g(x)=2x+7\]which is just a line

  7. KingGeorge
    • 4 years ago
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    You can only factor and cancel here because the denominator is equal to 0 at x=2. It's a little trick that follows from some theorems learned in pre-calc or calc 1. To show it has removable discontinuity at x=2, you only have to say that the only place the function is undefined, is at x=2 (obvious since you can't divide by 0), and show that a limit exists at x=2. To show the limit exists, you can check if the numerator, as well as the denominator is equal to 0 at x=2. If this is true, factor, cancel, and solve. Remember, this only works if the numerator AND the denominator are equal to 0.

  8. anonymous
    • 4 years ago
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    and the answer is that they are identical, except for the fact that f is not defined at 2, but g is

  9. anonymous
    • 4 years ago
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    so the graph of f looks just like the graph of g, a line with slope 2 and y intercept 7, except that at (2,11) where the function ought to be 11 there is a hole. other than that, they are identical

  10. anonymous
    • 4 years ago
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    meaning that if you define f to be 11 when x = 2 now it is in fact identical to the line \[y=2x+7\] and you have "removed" the discontinuity

  11. anonymous
    • 4 years ago
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    sorry it took me so long respond, this online work is being very slow... anyway say i have 2 equations \[(4x^3+24x^2+7x+42)/(x+6) and (2x^2+6x+a)\] the first formula has an x<-6 and the 2nd has x</=-6 and i have to find which value would make it continuous. i set them equal to each and plugged in -6 in for the x and solved for a getting -36 which is wrong

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