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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

You have:

x-9

----------

(√(x)+3)(x-9)

You cancel that out and get:

1

--------

(√(x)+3)

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

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