Limits with square roots

  • Thread starter subasurf
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  • #1
subasurf
1
0
Hi guys, I'm really new to calculus and limits and have been trying to have a good crack at the following question. Sorry if I haven't written the problem out in the most acceptable format.

lim (9-3√x)/(9-x)
x→9

Substituting 9 gives you 0/0 and indeterminate.

I tried multiplying the numerator and denominator by (-9+3√x), and multiplied out the brackets to wind up with -81/-27 and then simplified down to - 3/1

I'm just not sure if I'm going about this the correct way. I don't necessarily want the answer, just a bit of guidance. Thanks guys.
 

Answers and Replies

  • #2
clamtrox
938
9
I tried multiplying the numerator and denominator by (-9+3√x), and multiplied out the brackets to wind up with -81/-27 and then simplified down to - 3/1

There is some function f(x) that you can multiply numerator and denominator with, with the property that (9-3√x)f(x) = 9-x, but the function you are suggesting above is not it.
 

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