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anonymous
 4 years ago
can i get some help with limits?
anonymous
 4 years ago
can i get some help with limits?

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1Almost certainly. What kind of limits do you need help with?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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 \[(2x^2+3x14)/(x2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0check to see if 2 makes the numerator zero. if it does factor and cancel

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you get \[(x2)(2x+7)\] canceling gives you \[2x+7\] replace x by 2 and get 11

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't really get continuity at all i guess

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok the algebra is easy enough yes? so now the question is, what is the difference between \[f(x)=\frac{2x^2+3x14}{x2}\] and \[g(x)=2x+7\]which is just a line

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1You 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 precalc 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the answer is that they are identical, except for the fact that f is not defined at 2, but g is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0meaning 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry 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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