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How would you solve $\displaystyle\lim_{x\rightarrow-1}\left(\dfrac{\sqrt{x}-1}{x-1}\right)$ ?

I tried multiplying it by the conjugate. I don't know how to get rid of the square root.

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  • $\begingroup$ I tried multiplying it by the conjugate. I don't know how to get rid of the square root. $\endgroup$
    – Audrey
    Nov 11, 2014 at 10:07
  • $\begingroup$ You can't, it is out of the domain. $\endgroup$
    – user65203
    Nov 11, 2014 at 10:07
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    $\begingroup$ Are you sure that's $x \rightarrow -1$ and not $x \rightarrow 1^{-}$? $\endgroup$
    – daOnlyBG
    Nov 11, 2014 at 10:12
  • $\begingroup$ Yup, this is the question. I'm sure. $\endgroup$
    – Audrey
    Nov 11, 2014 at 10:15
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    $\begingroup$ Unless your course deals with complex numbers, $x\to-1$ is a typo for $x\to 1$. Assuming that it's a typo, you can use Autolatry' s hint. $\endgroup$ Nov 11, 2014 at 10:23

1 Answer 1

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Huge hint:

\begin{equation} (\sqrt{x}-1)(\sqrt{x}+1) = x-1 \end{equation}

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    $\begingroup$ But then it's gonna be 1/(x√+1). When you substitute x for -1, it's undefined $\endgroup$
    – Audrey
    Nov 11, 2014 at 10:15

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