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Almost certainly. What kind of limits do you need help with?
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 \[(2x^2+3x-14)/(x-2)\]
check to see if 2 makes the numerator zero. if it does factor and cancel
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\[(x-2)(2x+7)\] canceling gives you
\[2x+7\] replace x by 2 and get 11
i don't really get continuity at all i guess
ok the algebra is easy enough yes? so now the question is, what is the difference between
\[g(x)=2x+7\]which is just a line
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.
and the answer is that they are identical, except for the fact that f is not defined at 2, but g is
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
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
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