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
That should beOriginally 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.
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
Yes, but it is applicable only on rational or "fractional" functions. That is what L'Hopital's Rule is all about. Search the Internet for that rule so that you will understand why the limits of the derivatives of the numerator and denominator of a rational function will give the limit of that whole function.Originally Posted by tnkfub
L'Hopital or L'Hospital Rule.
Very good then. L'Hopital's Rule is really helpful in finding limits of rational functions when the limits are indeterminate. But that Rule is not always applicable, or will not always lead to the limit of the whole rational function. Later on you will hit the wall also in using that Rule, even after many differentiations. But for now, enjoy the Rule. I love it myself.Originally Posted by tnkfub