I got a couple questions coming, correcting a calculus test. I'll start with the limits I didn't get right: x -> 25, ((sqrt of x) - 5)/(x-25)

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I got a couple questions coming, correcting a calculus test. I'll start with the limits I didn't get right: x -> 25, ((sqrt of x) - 5)/(x-25)

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ok, this one's a little sneaky depending on how you look at it. you see the square root of x and 5 (which is the square root of 25) are in the numerator.
the denominator can then be factored via difference of squares:
(x-25) = (root[x]+ 5) (root[x]-5)

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Other answers:

indeterminate form , use L'hospital
root [x] - 5 appear on both the top and the bottom of the fraction, so you can then cancel both the terms out, leaving you with: \[\lim_{x \rightarrow 25} { 1 \over \sqrt{x} +5}\]
Man my teacher is an a-hole, he gives us very little time for practice than throws tricky questions like this at us. No matter how I looked at it I kept getting 0/0 which I knew couldn't be right. So it's 1/10. Sneaky indeed :D Thanks ggree, and might I say, nice pic! Adventure Time FTW!
The solution I posted for you is most likely the "correct" way for you, assuming you're in highschool or intro calculus. It might be worth it to learn l'hospital's rule, as roachie is suggesting. (note: your instructor might not accept it if you use it to solve a long answer problem. it will probably be ok to use on multiple choice if you've got it) Lhospital's rule is a seriously awesome shortcut if you use it right.
Duly noted, I've got about 15 questions to correct tonight, I'll look it up when I get the chance. Or even ask my teacher. Thanks for the clarification, I thought he was just talking gibberish :D
http://www.khanacademy.org/math/calculus/v/introduction-to-l-hopital-s-rule

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