Posted by Ezra on .
What is the limit (as u>2) of square root(4u+1)3/u2.
I got 1 as an answer, but I want to check if that's right. I heard it could be something else so I'm unsure and doubtful of my answer.

Calculus 
Steve,
take derivatives, so the limit is
2/√(4u+1) / 1 = 2/3 
Calculus 
Ezra,
I did it by multiplying by the conjugate. I'm not sure of how to do it any other way. Can you explain how that works?

Calculus 
Steve,
check your text for L'Hospital's (or L'HÃ´pital's) Rule. That is the usual tool used to evaluate limits when you have the form
0/0 or ∞/∞ or 0*∞ 
Calculus 
Reiny,
In most introductory Calculus courses, the study of limits precedes the concept of the derivative, since limits are used to develop the derivative by First Principles, so ...
using your idea of conjugates, .....
Lim ( √(4u+1)  3)/(u2) , u>2
= Lim ( √(4u+1)  3)/(u2) * (√(4u+1) + 3))/ (√(4u+1) + 3)) , u>2
= Lim ( 4u+1  9)/ ((√(4u+1) + 3))(u2))
= lim 4(u2)/ (√(4u+1) + 3))(x2)
= lim 4/ (√(4u+1) + 3)) , as u>2
= 4/6 = 2/3