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whats the limit of the following function?

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\[\frac{ x ^{2}-\sqrt{x} }{ 1-\sqrt{x} }\]
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Other answers:

you may try rationalizing, both numerator and denom separately..
but how can i do that? do first the denomitador and then de numerator or both at the same time?
doesnt matter..
any order you prefer..
The limit is -3.
you may rationalize either denom or nume first,,order wont matter..
Use L'Hospital Rule
mm but how? maybe something like this? \[\frac{ x ^{2} +\sqrt{x} }{ x ^{2} +\sqrt{x} }\] multiply the original function times the rationalization of the numerator?
i get 3x
yeep, i know the limit, but what i dont, is how to find it
Just take the derivative of the top, and then take the derivative of the bottom,
It is -3
Then plug in 1 for x and that will give you the limit.
mm but how can i find the limit without using L'hopital? cos my teacher doesnt allow me to do that yet, he want me to find the limit algebraically
Just multiply top and bottom by conjugates then.
respectively? i mean the numerator conjugate times the original numerator and the denominator conjugate times the original denominator?
What you want to do is rationalize one side.
So, just multiply top and bottom by 1+sqrt(x)
You should get 1-x for the bottom
\[\frac{ x ^{2}-\sqrt{x}(1+\sqrt{x}) }{1-x }\] so thts my result after doing what u said :p is it correct?
you would have parenthesis around x^2 and sqrt x
mm yes so what can i do next?
I am not sure actually you still have on the denominator 1-1 which is still dividing by a zero
your limit does not exist?
\(\huge\frac{ (x ^{2}-\sqrt{x})(1+\sqrt{x}) }{1-x }\times \frac{x^2+\sqrt x}{x^2+\sqrt x}\)
x^4-x = x(x^3-1) = x(x-1)(x^2x+x+1)
put x=1 after cancelling 1-x
divide both num and dem by x^2..

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