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I understand how to find limits, but for some reason I cannot figure out the algebra of this problem. I tried multiplying by the conjugate and end up with 0/0. When I check on my calculator, or apply L'Hopital's rule I get -1/6. Is there an algebra trick that I am missing on this one? $\displaystyle\frac{3-\sqrt{x+5}}{x-4}$

I have solved similar problems with the square root by multiplying by the conjugate, but it doesn't seem to work for this one.

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  • $\begingroup$ Note that $-x+4 = -(x-4)$ $\endgroup$ – Namaste Oct 19 '14 at 18:58
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\begin{gathered} \frac{{3 - \sqrt {x + 5} }}{{x - 4}} = \frac{{\left( {3 - \sqrt {x + 5} } \right)\left( {3 + \sqrt {x + 5} } \right)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ = \frac{{{3^2} - {{\sqrt {x + 5} }^2}}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ = \frac{{9 - (x + 5)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ = \frac{{ - (x - 4)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\ = \frac{{ - 1}}{{\left( {3 + \sqrt {x + 5} } \right)}} \\ \end{gathered}

Take the limit of that as $x$ approaches $4$.

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  • $\begingroup$ Thank you! It was the step where 9-x-5 becomes -1(x-4) that I was missing! $\endgroup$ – mathtimefun Oct 19 '14 at 19:13
  • $\begingroup$ @mathtimefun: You are welcome! $\endgroup$ – Rory Daulton Oct 19 '14 at 22:46
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When you multiply with that conjugate, you end up with $-x+4$ in the NUM. So that "cancels" against that x-4 in the DENOM (apart from that minus) Retaking the limit gives that -1/6

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$$\begin{array}{rcl}\lim_{x\to 4} \frac{3-\sqrt{x+5}}{x-4} & = & \lim_{x\to 4} \frac{(3-\sqrt{x+5})(3+\sqrt{x+5})}{(x-4)(3+\sqrt{x+5})}=\lim_{x\to 4} \frac{9-(x+5)}{(x-4)(3+\sqrt{x+5})} \\ & = & \lim_{x\to 4} \frac{4-x}{(x-4)(3+\sqrt{x+5})} =\lim_{x\to 4} \frac{-1}{3+\sqrt{x+5}}=- \frac{1}{6}.\end{array}$$

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