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anonymous
 one year ago
I would like to know if I am understanding the concept of limits and derivatives
limx→1 (x^2+3x−4) / (x^2+8x−9) =(limx→1(x^2+3x−4)) / (limx→1 (x^2+8x−9))
This is false because the first part equals 1/2 and the second part is indeterminate
If f′(2) exists, then then the limit limx→2f(x) is f(2)
This is true because there is continuity.
If limx→3f(x)=∞ and limx→3g(x)=∞, then limx→3[f(x)−g(x)]=0
This is true because of the limit law of quotient.
If limx→2[f(x)g(x)] exists, then the limit is f(2)g(2)
This is true because of the limit law of product.
anonymous
 one year ago
I would like to know if I am understanding the concept of limits and derivatives limx→1 (x^2+3x−4) / (x^2+8x−9) =(limx→1(x^2+3x−4)) / (limx→1 (x^2+8x−9)) This is false because the first part equals 1/2 and the second part is indeterminate If f′(2) exists, then then the limit limx→2f(x) is f(2) This is true because there is continuity. If limx→3f(x)=∞ and limx→3g(x)=∞, then limx→3[f(x)−g(x)]=0 This is true because of the limit law of quotient. If limx→2[f(x)g(x)] exists, then the limit is f(2)g(2) This is true because of the limit law of product.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2`If limx→3f(x)=∞ and limx→3g(x)=∞, then limx→3[f(x)−g(x)]=0 ` `This is true because of the limit law of quotient.` I don't agree. Recall that one of the many indeterminant forms is \(\Large \infty  \infty\)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2sorry it's spelled "indeterminate"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops, I meant the limit law of difference not quotient. However, I now recalled that indeterminate form. So that statement would be entirely false because it is indeterminate?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah `limx→3[f(x)−g(x)]` is indeterminate and not equal to 0
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