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I've the following limit to be calculated.

$\lim_{x \to 7} \dfrac{5 - \sqrt{4 + 3x}}{7 - x}$

I'm used to solve this limits from Calculus I text books but I've been struggling without success on this one, how make it non-indeterminable ?

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  • $\begingroup$ sorry, I had a typo, fixed now. $\endgroup$
    – aajjbb
    Sep 10 '14 at 21:04
  • $\begingroup$ Now the numerator isn't a real number near $x=7$. Do you really mean $x\rightarrow -7$? $\endgroup$
    – MPW
    Sep 10 '14 at 21:05
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    $\begingroup$ I suspect the radicand should be $4 + 3x$, giving $\to \frac 00$ $\endgroup$
    – amWhy
    Sep 10 '14 at 21:06
  • $\begingroup$ Once OP fixes the problem, he might consider L'Hopital $\endgroup$
    – MPW
    Sep 10 '14 at 21:08
  • $\begingroup$ yes @amWhy, I had another typo, sorry. $\endgroup$
    – aajjbb
    Sep 10 '14 at 21:09
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Hint: multiply by $$\frac{5+\sqrt{4+3x}}{5+\sqrt{4+3x}}$$ And recall that $(\alpha-\beta)(\alpha+\beta) = \alpha^2 - \beta^2$. This trick is called multiplying by the conjugate, and it is a way of getting rid of those roots and the $\frac{0}{0}$ type of indetermination.

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  • $\begingroup$ You need + under radical ... OP edited problem. :) $\endgroup$ Sep 10 '14 at 21:19
  • $\begingroup$ You're correct indeed, wrote that right in my whiteboard though. :) $\endgroup$ Sep 10 '14 at 21:19
  • $\begingroup$ With this multiplication, the upper side will become 0. $\endgroup$
    – aajjbb
    Sep 10 '14 at 21:23
  • $\begingroup$ @aajjbb How come?, the numerator goes like this: $(5-\sqrt{4+3x})(5+\sqrt{4+3x}) = 5^2 - (\sqrt{4+3x})^2 = 25 - (4+3x) = 21 - 3x = 3(7-x)$. $\endgroup$ Sep 10 '14 at 21:25
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    $\begingroup$ noticed it. Thank you. $\endgroup$
    – aajjbb
    Sep 10 '14 at 21:34
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By L'Hopital theorem we have $$\lim\limits_{x\to7}\frac{5-\sqrt{4+3x}}{7-x}=\lim\limits_{x\to7}\frac{-3}{-2\sqrt{4+3x}}=\frac{3}{10}$$

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