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Take the limit as x approaches 9 of $\frac{3}{\sqrt{x}-3} - \frac{18}{x-9}$ Multiplying by the conjugate $\sqrt{x}+3$ on the left side to get a common denominator of $x-9$, then using L'Hôpital's rule and solving gets the answer 1/2 Multiplying the denominators together to get a common denominator then using L'Hôpital's rule and solving get me 3 on the numerator and 0 on the denominator. I've done this problem a few times now but I cannot seem to find my mistake.

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  • $\begingroup$ Hi and welcome to Math.SE! Please show your second calculation in more detail. It must be wrong somewhere, but since you don't show it to us, how should we know where? Also, try to format your question with mathjax, it will be easier for others to read. I noted you just got a down vote. It might be for not showing your calculations... $\endgroup$ – mickep Sep 13 '15 at 7:36
  • $\begingroup$ Or at least post a picture of your equations. $\endgroup$ – Antitheos Sep 13 '15 at 7:40
  • $\begingroup$ You should do the calculation really step by step. The solution to the problem should be very clear and specific. The more information we have, the better we can help you. You should also get used to mathJax, (If you are not already) in order to write equations etc. $\endgroup$ – Imago Sep 13 '15 at 7:48
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I am getting $1/2$ in both cases.

1st method: Applying L'Hôpital Rule (I have eaten the steps involved and shown only 1st equation and last equation)

$$\lim_{x\to9} \left(\frac{3}{\sqrt{x}-3}-\frac{18}{x-9}\right)=\lim_{x\to9} \frac{3}{2\sqrt{x}}=\frac{1}{2}.$$

2nd method: Applying L'Hôpital Rule (I have again eaten the steps involved and shown only 1st equation and last equation)

$$\lim_{x\to9} \frac{3x-18\sqrt{x}+27}{x^{3/2}-3x-9\sqrt{x}+27}=\lim_{x\to9}\frac{1}{\sqrt{x}-1}=\frac{1}{2}.$$

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  • $\begingroup$ Can you please give me the link to that tutorial? Thanks $\endgroup$ – dknight Sep 13 '15 at 8:02
  • $\begingroup$ @gaurav I've edited your answer by formating and re-ordering some text. Let me know if it is what you wanted to say. $\endgroup$ – Nicolas Sep 13 '15 at 8:04
  • $\begingroup$ @Nicolas yes thats exactly what I wanted, can you also give me the link to the tutorial Morgan Rodgers was referring to? Thanks $\endgroup$ – dknight Sep 13 '15 at 8:06
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    $\begingroup$ Here you go: MathJax basic tutorial and quick reference $\endgroup$ – rubik Sep 13 '15 at 8:07
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I made a mistake, my numerator was actually 0 not 3, which would result in "0/0" again. I would then have to apply l'hôpital's rule again then solve. My bad

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    $\begingroup$ In this case I would delete the question, since there was no actual problem. $\endgroup$ – rubik Sep 13 '15 at 8:05

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