Limit At Infinity For Square Root Expression Horizontal Asymptote TIPS 5

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  1. Anil Kumar
  2. learning
  3. centre
  4. aan
  5. AAN Learning Centre
  6. square root
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Комментарии
Автор

I am in the alleged best university in my country and you just made more sense in the first five minutes than my lecturer ever has

cuisinepersan
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If we extend the reals to include infinitie numbers and let infinity be omega (the number of all natural numbers), we can calculate the result of b. = 2 omega - 0.5.

MisterrLi
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what is the 1/x^2 actually supposed to be? should it be x/x^2 if the proper conjugate is used? in this case would the limit still be 1/2? i have my midterm tonight... :( thanks :)

brianportner
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you messed up in the first step when you multiply by the conjugate the numerator and denominator do not match buddy

UaaaGlobal
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Everything except the conjugate is correct.

toxi
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its wrong. u multiplied with wrong conjugate

rashad
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Why √x^2+1 +x, why 1??? Shouldn't it be X????

myahridurdynazarova