Question

What is the limit of ((x^.5)-3)/(x-9) as x approaches 9?

Answer 1

I'm not sure if I should use l'Hopital's or what

Answer 2

\[\frac{ x ^{5}-3 }{ x-9}\] correct?

Answer 3

No, it's x^.5, not x^5. So, the square root.

Answer 4

Oh, square root. Alrighty then. Well l'hopitals rule is valid when you get an indeterminant form like 0/0. So since you get that, l'hopitals is perfectly valid. You know how to use it?

Answer 5

Yeah, I used it once and it still didn't work so I figured I should check if I was doing the right thing before I did f''. Sort of messy haha.

Answer 6

Alright, let me check then : )

Answer 7

I will too, thanks a bunch by the way.

Answer 8

Shouldnt be ugly at all :P

Answer 9

Wait, since it'll always be over the denominator raised to some power, won't it always be divided by zero?

Answer 10

Nope. So we need to take the derivative of the numerator and the derivative of the denominator. We do these separately, not together. So the denominator derivative will just be 1. As for the numerator:

\[\sqrt{x}-3 = \frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }} = \frac{ 1 }{ 2\sqrt{x} }\]
So since the denominator derivative became 1, all you have remaining is the derivative I just did. Plug in 9 and you got your answer :3

Answer 11

Why is it that you don't use the quotient rule, since you're finding the derivative of a "fraction" equation?

Answer 12

I can understand how that would be confusing when they state the rule to you. Even though they say it is f'(x)/g'(x), they DO NOT mean that it is quotient rule. They want you to treat the numerator as a separate function and the denominator as a separate function.

Answer 13

Okay! Thanks so much.:)

Answer 14

Yeah, people get that confused xD But hope it makes sense then :3