A limit equation :How to prove -------

[v9260019]

how to prove a limit

Hello all.
I have a question about limit question:
How to prove: limit[x->Infinite]{Sqrt[x]-ln[x]}=Infinite
Thanks a lot

 

[treehugger]

Hello all.
I have a question about limit question:
How to prove: limit[x->Infinite]{Sqrt[x]-ln[x]}=Infinite
Thanks a lot

instead of using every value on the function lets sample it using x^2. this is like chosing a sub-series out of the actual series.

now we have lim(x->infinite) {Sqrt(x^2) - ln(x^2)} = lim(x->infinite){x - 2lnx}

since derivative of x - 2lnx converges to 1, lim(x->infinite){x-2lnx}=infinite.

since a subseries of the actual series diverges, our original series should also diverge.

 

[steve10]

To prove limit[x->Infinite]{f(x)}=Infinite for a function f(x). Here is what you should do:

For any M>0, you are required to find an N>0, such that, when x>N, f(x)>M.

For your function f(x)=Sqrt[x]-ln[x], write it in the folloiwng form:
f(x) = Sqrt[x]/2 + (Sqrt[x]/2-ln[x])

First, choose an N1>0 such that, when x>N1, the second term of the right hand side (Sqrt[x]/2-ln[x])
>=0. Next, choose an N2>0 such that, when x>N2, the first term of the right hand side Sqrt[x]/2 >M. Then when x>max{N1,N2}, f(x)>M.

In the above, the only work is at choosing N1, which would take you a little effort. All others are trivial.

 

[mehtesham]

one strange way for this kind of problems is to compare √x and ln(x) when
x->infinite. this is done by clac. lim(√x/ln(x),x->∞)=∞ or
lim(ln(x)/√x,x->∞)=0.these shows √x is infinite bigger than ln(x) when
x->infinite then lim(√x-ln(x),x->∞)=∞.

 

[teteamigo]

Too simple...

Put it in the form:

lim{x->+00} [ ln(e^(sqrt(x)) -ln(x) ] = lim{x->+00} ln( e^(sqrt(x) ) / x) =*


e^(sqrt(x) ) / x is continuous in ]0,+00[ then lim {x->+00} e^(sqrt(x) ) / x= e^(sqrt(x0) ) / x0


*=lim {x->+00} ln( lim {x->+00} e^(sqrt(x) ) / x ) =
lim {x->+00} ( +00)=+00