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I need to prove the limit as x approaches 1 of x^(1/2)=1. In order to i have have to figure out how to factor x^(1/2)-1. I am stuck at factoring it.
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why?
i mean why do you have to factor?
to prove that lx^(1/2)-1l< eplsilon when delta>lx-1l. \[\to prove that \left| \sqrt{x}-1 \right|<\epsilon when \left| x-1 \right|< \]
you could try saying, for example, that since x is approaching 1, it is certainly less than 4 for \(\sqrt{x}+1<3\) giving you \[|\sqrt{x}-1|<\epsilon \] \[|\sqrt{x}-1||\sqrt{x}+1|<3\epsilon\] \[|x-1|<3\epsilon\] so you can pick a \(\delta=\frac{\epsilon}{3}\) and run the argument backwards as usual
I will try that thank you
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