Calculus: limits
posted by Kay .
Evaluate the limit at t > 25: [25  t] / [5  √(t)]
I multiplied it by √(x + 3) to get rid of the square root on the bottom. However, once I'm done distributing it out, I'm not sure I'm doing it right.
I get [x√(x+3) + 2x  2] / [x + 5]
Once I plug in 25 for t, I get really big numbers and I end up getting an off number that is not correct (like 25√28 + 48 / 30)
Can someone please help?

Calculus: limits 
Ian
Factor (25t) into (5sqrt(t)) and (5+sqrt(t))
Cancel top and bottom (5sqrt(t) and left with lim t>25 of (5+sqrt(t))
which is 10. 
Calculus: limits 
MathMate
The idea of rationalizing the denominator is excellent.
You may note that the denominator is of the form (AB), so if you multiply top and bottom by (A+B), you will get (A²B²) in the denominator.
Since you are taking the limit, t does not equal 25, so you are allowed to divide top and bottom by (25t). Simplify and get a neat answer in no time.
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