limits
you have multiply by the conjugate.
to*
okay sooooo the limit as x approaches 25 when f(x)=(25-x)/(sqrt(x)-5) understand that you can try looking at (25-x) as (a^2-b^2) since (a^2-b^2) can be factored into (a-b)(a+b), you can make (25-x) into (5-sqrt(x))(5+sqrt(x))
Now we have (5-sqrt(x))(5+sqrt(x)) ----------------- sqrt(x)-5
Then you can factor out a -1 from the denominator to get: (-1) (sqrt(x)-5)
(-1) (5-sqrt(x)) excuse me
With that as a denominator, you'll see that our equation is now (5-sqrt(x))(5+sqrt(x)) ----------------- (-1)(5-sqrt(x))
Since both the numerator and denominator have (5-sqrt(x)), that can be cancelled out
Got the rest on your own? :)
is it 1?
Well once you cancel out the (5-sqrt(x)), you're left with (5+sqrt(x)) -------- -1 Since you are looking for the limit as x approaches 25, you simply plug 25 into x
5+ sqrt(25) -------- -1
What do you get for that?
-10 (:
There you go! *high five* lol :)
thanks!!!
anytime :)
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