1. limit problem

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2. Originally Posted by Revelsyn
lim u --> 1 (sqrt1+u) - (sqrt1+u^2)/ (u-1)

i tried to solve this many times without success, and i have a feeling it's very obvious ><

do you mean $\frac {\sqrt{1 + u} - \sqrt{1 + u^2}}{u - 1}$ ?

if so, multiply by $\frac {\sqrt{1 + u} + \sqrt{1 + u^2}}{\sqrt{1 + u} + \sqrt{1 + u^2}}$ (the conjugate of the numerator over itself) and simplify as far as possible before taking the limit. you should be able to cancel out the denominator and hence remove the problem

3. Rationalize the numerator:
$\lim_{u \to 1} \left( \frac{\sqrt{1+u} - \sqrt{1+u^2}}{u-1} \cdot \frac{{\color{red}\sqrt{1+u} + \sqrt{1+u^2}}}{{\color{red}\sqrt{1+u} + \sqrt{1+u^2}}}\right)$

Simplify the numerator (notice it is a difference of squares and don't worry about the denominator .. you'll see why)

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5. Originally Posted by Revelsyn
then i did :
=(1+ u) - (1 + u^2)/(u-1)[(sqrt1+u) + (sqrt1+u^2)]

=u - u^2/ (u-1)[(sqrt1+u) + (sqrt1+u^2)]

=-u(u-1)/(u-1)[(sqrt1+u) + (sqrt1+u^2)]

=-1/2(sqrt2) ???
yes

you are correct

(note that you have to rationalize the fraction to make the answer look like the one you want)

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