anonymous
  • anonymous
Help- find the limit of:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ x^2+2x-8 } \right)+\left(\frac{ x-6 }{ x^2-4} \right)\]
anonymous
  • anonymous
so I found that \[\lim_{x \rightarrow 2} \left( \frac{ 6 }{ (x+4)(x-2) } \right)+\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\] Now what?????
nubeer
  • nubeer
now subsitute 2 in place of x

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More answers

anonymous
  • anonymous
it will be 6/0. error. I can't do that yet.
nubeer
  • nubeer
1/0 = infinity
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\lim_{x \rightarrow 2}\left( \frac{ 6(x+2)+(x-6)(x+4) }{ (x+4)(x-2)(x+2) } \right)\]
nubeer
  • nubeer
they took the LCM in this step
anonymous
  • anonymous
ok. Can you explain that part to me? I don't seem to get it.
nubeer
  • nubeer
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 (x-2) is written twice so u will just write it one time
anonymous
  • anonymous
ok but what about the numerator? How did they get to that part?
nubeer
  • nubeer
ok now move back a step and see see what u had in denominator before . u had (x+4)(x−2) right?
nubeer
  • nubeer
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
nubeer
  • nubeer
|dw:1353067137453:dw|
nubeer
  • nubeer
|dw:1353067229340:dw|
nubeer
  • nubeer
|dw:1353067272848:dw|
nubeer
  • nubeer
|dw:1353067293846:dw|
nubeer
  • nubeer
are you getting what i am doing?
anonymous
  • anonymous
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.
Zarkon
  • Zarkon
\[\lim_{x \rightarrow 2} \left[\left( \frac{ 6 }{ (x+4)(x-2) } \right)+\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\right]\] \[\lim_{x \rightarrow 2} \left[\frac{x+2}{x+2}\left( \frac{ 6 }{ (x+4)(x-2) } \right)+\frac{x+4}{x+4}\left( \frac{ x-6 }{ (x+2)(x-2) } \right)\right]\]
anonymous
  • anonymous
thnk

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