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anonymous
 5 years ago
evaluate lim (x^21)/(sqrt(x)1)
x>
the answer in the book says 2
anonymous
 5 years ago
evaluate lim (x^21)/(sqrt(x)1) x> the answer in the book says 2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lim x>1 x^21/ sqrt(x) 1 = lim x>1 ( sqrt(x) 1 )( sqrt(x) +1 ) / ( sqrt(x) 1 ) = lim x>1 ( sqrt(x) +1) pluggin in the values , we get = ( sqrt(1) +1) = 1 +1 =2 hope that helps !!! let me know if u r satisfied...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I dont understand the first step

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember the formula a^2  b^2 = (a  b) (a +b)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(sqrt(x)1) (sqrt(x)+1) does that equal x1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u got it ...cheers !!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the original problem has a numerator of x^21 not X1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the square root of x^2 is only x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay !!! then make it as (x 1)(x +1) which equals to (sqrt(x)1) (sqrt(x)+1) (x +1) and now plug in the values

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but thats like we are getting 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought this one looked so simple

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r u sure that u read the statement correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0evaluate the indicated limit, if it exists. lol (x1)/(sqrt(x)1) I was wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes.......the limit does exist which is 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your original answer is right, Thank you very much for your help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02 because I had the problem wrong the numerator is (x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh !!!!!! finally we r there !!! gud luck with ur maths ..
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