If you can help me on this id greatly appreciate it

(steps in helping me understand helps alot)

Lim ((1/x^(1/2))-(1/2))/(x-4)

x->4

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- Feb 7th 2006, 09:14 PMtnkfublimit help
If you can help me on this id greatly appreciate it

(steps in helping me understand helps alot)

Lim ((1/x^(1/2))-(1/2))/(x-4)

x->4 - Feb 8th 2006, 12:04 AMticbolQuote:

Originally Posted by**tnkfub**

Lim(x-->4)[(1/sqrt(x) -1/2) /(x-4)] because you're on L'Hopital's Rule, I guess.

So, substitute 4 for all the x's,

= (1/sqrt(4) -1/2) /(4-4)

= (1/2 -1/2) /0

= 0/0

Indeterminate.

Use the L'Hopital's Rule in this 0/0 limit case.

Lim(x->a)[f(x)/g(x)] = Lim(x->a)[f'(x)/g'(x)] ----***

f(x) = 1/sqrt(x) -1/2 = (x)^(-1/2) -1/2

f'(x) = (-1/2)*x^(-3/2) = -1/[2sqrt(x^3)] ----***

g(x) = x -4

g'(x) = 1

So,

Lim(x->4)[(1/x^(1/2) -1/2) /(x-4)]

= 0/0

= Lim(x->4)[(-1/[2sqrt(x^3)) /(1)]

= -1/[2sqrt(4^3)] /1

= -1/[2sqrt(64)]

= -1/[2*8]

= -1/16 -----------answer. - Feb 8th 2006, 09:26 AMtnkfubThanks + Question :P
hey thanks alot for helping me on that but i have one question and if this is a yes than this makes lim like this alot easier >>> Can you use the derivative of limits and get the same answer?

thats what you did right use the derivative to come up with that answer - Feb 8th 2006, 10:05 AMticbolQuote:

Originally Posted by**tnkfub**

L'Hopital or L'Hospital Rule. - Feb 8th 2006, 09:48 PMtnkfub
thanks alot uve helped me even more than by solving that problem for me :P cuz i also have a test today on this material and this L'Hospital Rule is gonna be real useful :P

- Feb 8th 2006, 10:10 PMticbolQuote:

Originally Posted by**tnkfub**