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anonymous
 5 years ago
Evaluate the limit, if it exist
lim as x approaches 1, (x^31)/(x^21).
anonymous
 5 years ago
Evaluate the limit, if it exist lim as x approaches 1, (x^31)/(x^21).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please show step by step thanks

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it the general hospital rule...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok factor top and bottom \[\frac{x^31}{x^21}=\frac{(x1)(x^2+x+1)}{(x+1)(x1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now cancel the x  1 to get \[\frac{x^2+x+1}{x+1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0replace x by 1, get \[\frac{3}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oooooh ok i tried the other way and thought that you could factor it out...im still trying to understand where the x^2+x +1 came from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this kind of problem tends to come before differentiation and therefore before l'hopital's rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0difference of two cubes! \[a^3b^3=(ab)(a^2+ab+b^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in your case a = x and b = 1 to give \[x^31=(x1)(x^2+x+1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically you were factoring it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok cool i knew that what the steps were...well thanks again for the help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btw if you replace x by 1 in the numerator you get 0 yes? this means the numerator MUST factor as (x1) times something.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea.. thats what i did but somehow did it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is not a miracle you can factor, it is a consequence of the fact that you get 0 if you replace x by 1. same as in the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea thats exactly what i did so i figure theres something more to it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks again your a life saver
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