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Advanced Math/[x-9][sqrt(x)+3]


Over the summer, the AP calculus teacher at my school gives the incoming students a packet of problems he wants us to look up and be familiar with completing before the new school year starts. One of the questions was to find the limit of [sqrt(x)-3]/[x-9] as x approaches 9. After plugging in the 9 and getting 0/0, I multiplied the original function by the conjugate binomial of the numerator. After doing so, I got (x-9) for the numerator, but I'm not entirely sure how to multiply [x-9] by [sqrt(x)+3]... If you could offer any assistance that would be great! Thanks

Don't. You leave the denominator as it is. Why would you multiply it out.

If the denominator is (√(x)+3)(x-9), then it is already factored . If you were to multiply it out, the next step in your mind would probably be "oh, can I factor this?" which is just to undo the process of multiplying it out.

You have:


You cancel that out and get:


And that's it. To take the limit, plug in 9 and get 1/6.

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Clyde Oliver


I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.


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