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clara1223
 one year ago
find the limit as h approaches 0 of h(3+(2/h^2))
clara1223
 one year ago
find the limit as h approaches 0 of h(3+(2/h^2))

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clara1223
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0} h(3+\frac{ 2 }{ h ^{2} })\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we can try some test points, say h=.0000001, and h=.0000001 but what does your thoughts tell you initially?

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0Well I solved and got that the limit is 2, but that was wrong.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1your solution was flawed then :) we have 2 terms: ab if h < 0, a=, b=+ ... ab =  if h > 0, a=+,b=+ ... ab=+

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1let a=h let b=3+2/h^2 then multiply them together and yo have your setup. when h is less than zero, what are our signs for a and b? what about their product?

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0if h is less than 0 then a would be negative and b would be positive, and their products would both be negative

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ok, so we have a left side limit that is negative ... now assume h is greater than zero ... what are the signs of a and b? and of their product?

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0if h is greater than 0 then a and b would both be positive and so would be their products.

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0oh, so since from the left it's negative and from the right it's positive, it doesnt exist?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so we have a right side limit that is positive can a positive limit ever be equal to a negative limit?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1correct, its actually inf and +inf so yeah

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0yeah, I graphed it and it helped me to understand.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1graphs do help if applicable :)
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