anonymous
  • anonymous
Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
By using l'hopital's rule, take the derivative of \[x ^{2}+x = 2x+1\] and the derivative of \[x+1 = 1\] \[A = +\infty\] Similary for B = 1
anonymous
  • anonymous
im not familiar with the l'hopital rule
anonymous
  • anonymous
is there a way i can do it geometrically

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anonymous
  • anonymous
I don't think so. L'hopital's rule states that if you plug in the infinity into that equation and when the equations gives \[\infty/\infty\] or 0/0 you take the derivative of the equations separetly.
anonymous
  • anonymous
since im not familiar with this rule, can you walk me through it
anonymous
  • anonymous
Did you read my post above? I gave you a little info about the rule. If you need detailed help just tell me.
anonymous
  • anonymous
yes please, i need detailed help
anonymous
  • anonymous
For example take the equation above \[A = x ^{2}+x/x+1\] If you replace +infinity in the all of the x-s you get infinity/infinity which is an indeterminate form i.e. you cannot compute the answer. Because it is an indeterminate form you have to take the derivative of those equation seperatly, like i did in my first post.
anonymous
  • anonymous
ok so what happens to the 2x+1/1
anonymous
  • anonymous
replace infinity in that x which gives\[2(\infty)+1/1 = \infty\]
anonymous
  • anonymous
oh ok i see
anonymous
  • anonymous
so B= 2x+1/2x+1
anonymous
  • anonymous
2(-inf)+1/2(-inf)+1 = -inf?
anonymous
  • anonymous
No. The derivative of 1 = 0. The derivative of any constant is equal to 0
anonymous
  • anonymous
i understand now thanks

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