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Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ x^2+2x8 } \right)+\left(\frac{ x6 }{ x^24} \right)\]
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
so I found that \[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ (x+4)(x2) } \right)+\left( \frac{ x6 }{ (x+2)(x2) } \right)\] Now what?????
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
now subsitute 2 in place of x
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
it will be 6/0. error. I can't do that yet.
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
1/0 = infinity
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
sorry 1/3. I see the next step in the book but I don't understand how they got to it. I will write it up.
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 2}\left( \frac{ 6(x+2)+(x6)(x+4) }{ (x+4)(x2)(x+2) } \right)\]
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
they took the LCM in this step
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
ok. Can you explain that part to me? I don't seem to get it.
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
write all the terms in denominator if some term is repeating more then once , just write it one time. like in this case on one side u have (x+4)(x−2) and on other side (x+2)(x−2) so write them all in denominator but (x2) is written twice so u will just write it one time
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
ok but what about the numerator? How did they get to that part?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
ok now move back a step and see see what u had in denominator before . u had (x+4)(x−2) right?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
so you compare the old denominator with new numerator whichever is the common denominator u cancel them and whatever is left u multiply it with numerator
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
dw:1353067137453:dw
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
dw:1353067229340:dw
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
dw:1353067272848:dw
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
dw:1353067293846:dw
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.1
are you getting what i am doing?
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
yes, so basically you devide the entire equation the the common denoiminator and then solve from there. I think I got it. Let me see if it works out.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 2} \left[\left( \frac{ 6 }{ (x+4)(x2) } \right)+\left( \frac{ x6 }{ (x+2)(x2) } \right)\right]\] \[\lim_{x \rightarrow 2} \left[\frac{x+2}{x+2}\left( \frac{ 6 }{ (x+4)(x2) } \right)+\frac{x+4}{x+4}\left( \frac{ x6 }{ (x+2)(x2) } \right)\right]\]
 one year ago

Compgroupmail Group TitleBest ResponseYou've already chosen the best response.0
thnk
 one year ago
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