Question

Use the Mean Value Theorem to show that for any n>0
\(\dfrac{n}{2n+1}<\sqrt{n^2+n}-n<\dfrac{1}{2}\)
Please, help

Answer 1

i have no idea how to do this, but i do have a suggestion
use the mvt only for the first inequality
the second follows from the fact that
\[\lim_{n\to \infty}\sqrt{n^2+n}-n=\frac{1}{2}\]

Answer 2

Apply the MVT to \(f(x)=\sqrt{n^2+x}\) for \(0\le x\le n\)

Answer 3

@Zarkon I got the right hand side, but still stuck at the left hand side
Here is my work:
1) f(x) is continuous on [0,n]
2) \(f'(x) = \dfrac{1}{2\sqrt{n^2+x}}\) exists for all x on [0,n]
hence MVT gives us \(f'(c) = \dfrac{f(n)-f(0)}{n-0}=\dfrac{\sqrt{n^2+n}-\sqrt{n^2}}{n}=\dfrac{\sqrt{n^2+n}-n}{n}\)

Answer 4

Moreover, we have \(f'(c) =\dfrac{1}{2\sqrt{n^2+c}}\) hence
\(\dfrac{1}{2\sqrt{n^2+c}}=\dfrac{\sqrt{n^2+n}-n}{n}\)

Answer 5

That is \(\sqrt{n^2+n}-n=\dfrac{n}{2\sqrt{n^2+c}}\)

Answer 6

The right hand side : \(\dfrac{n}{2\sqrt{n^2+c}}=\dfrac{n}{2n\sqrt{1+c/n^2}}=\dfrac{1}{2\sqrt{1+c/n^2}} < \dfrac{1}{2}\)

Answer 7

hence \(\sqrt{n^2+n}-n < \dfrac{1}{2} \) for n > 0

Answer 8

oh, I got the left hand side also. Thanks a tooooon . lalala...

Answer 9

something is wrong i think the MVT should be applied to
\[\large \rm f(x)=\sqrt{x^2+x}\] for \[\rm 0<c<x\]
?

Answer 10

hmm i still have to go over it heheh
i thought there was an error haha