Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$

Question

Prove that

$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$

$$\text{if and only if}$$ $$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck $$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$ Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$

Edit: x tends to infinity, not to 0. I transcribed wrongly.

Please edit this into the title; I clicked on the question knowing that the statement is simply false as written. — , Jun 15 '16 at 14:52

score 6 · Answer 1 · answered Jun 15 '16 at 13:36

6

This is simply false.
As $\lim_{x \to 0} x + n = n$ for all $n \in \mathbb{N}$ and the square root is continuous, we have $\lim_{x \to 0} \sqrt{x+n} = \sqrt{n}$.
Then, you must show \begin{align*} a + \sqrt{2} b + \sqrt{3}c = 0 \iff a+b+c =0. \end{align*} Take $a = 2$, $b = -\sqrt{2}$ and $c=0$, then the LHS is satisfied, but the RHS is not.
With $a=1$, $b=-1$ and $c=0$, the LHS is not satisfied, but the RHS is.

answered Jun 15 '16 at 13:36

Hetebrij

3,981

Perhaps, but why should they be integers? $x$ is clearly not an integer. – Hetebrij Jun 15 '16 at 13:41
The statement would still be false: Take $a=2,b=c=-1$. – Micapps Jun 15 '16 at 13:42
@Micapps, but it would be less false, if the limit is zero, then $a+b+c=0$, so one direction holds. – Hetebrij Jun 15 '16 at 13:44
I have o ploted the function in Desmos with a and b parameters and c=-a-b so that a+b+c=0, for any a and b in (-10, 10), the functions tends to 0. – Felix Mada Jun 15 '16 at 13:47
@Hetebrij I guess... How much energy should we expend debugging this? If $a,b,c$ are integers then the least false statement is probably that the limit =0 iff $a=b=c=0$. – Micapps Jun 15 '16 at 13:47
@FelixMada Tending to zero doesn`t equal 0. – Hetebrij Jun 15 '16 at 13:53
@FelixMada, I also graphed in desmos using $a = 1, b = -1, c=0$ (one of the sets of values given in this answer) and the function does not tend to 0 as $x \to 0.$ – Jun 15 '16 at 13:58
It tends to $1- \sqrt{2}$ as we would expect algebraically. – Jun 15 '16 at 13:59
You say $x \to 0$, the OP says $x \to \infty$. – marty cohen Jun 16 '16 at 19:57
Um, x -> infinity not 0. root(x + k) and root(x) both tend to infinity. Not root k and 0. – fleablood Jun 16 '16 at 20:49

score 2 · Answer 2 · answered Jun 16 '16 at 19:18

For all $x>0$ one has $$\eqalign{\sum_{i=1}^3 a_i\sqrt{x+i}&=\sum_{i=1}^3 a_i\bigl(\sqrt{x+i}-\sqrt{x}\bigr)+\sqrt{x}\sum_{i=1}^3 a_i\cr &=\sum_{i=1}^3 {ia_i\over\sqrt{x+i}+\sqrt{x}}+\sqrt{x}\sum_{i=1}^3 a_i\ .\cr}.$$ Now let $x\to \infty$, and the claim is immediate.

score 0 · Answer 3 · answered Jun 15 '16 at 14:57

Note that we have

$$\begin{align} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}&=\sqrt{x}\left(a\sqrt{1+\frac{1}{x}}+b\sqrt{1+\frac{2}{x}}+c\sqrt{1+\frac{3}{x}}\right)\\\\ &=a\sqrt{x}\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\\\\ &+b\sqrt{x}\left(1+\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\\\\ &+c\sqrt{x}\left(1+\frac{3}{2x}+O\left(\frac{1}{x^2}\right)\right)\\\\ &=(a+b+c)\sqrt{x}+\frac{(a+2b+3c)}{2\sqrt{x}}+O\left(\frac{1}{x^{3/2}}\right) \tag 1 \end{align}$$

from which we see that the limit is zero if and only if $a+b+c=0$

Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark — Mark Viola, Jun 22 '16 at 17:46

score 0 · Answer 4 · answered Jun 16 '16 at 19:14

The limit is bounded if a+b+c = 0

verify: $\lim_\limits{x \to \infty} \sqrt{x+k} - \sqrt {x} = 0$

$\lim_\limits{x \to \infty} a\sqrt{x+1} -a\sqrt{x}+ b\sqrt{x+2} -b\sqrt{x} + c\sqrt{x+3} -c\sqrt{x}= 0$

$\lim_\limits{x \to \infty} (a\sqrt{x+1} + b\sqrt{x+2} + c\sqrt{x+3}) = \lim_\limits{x \to \infty}(a+b+c) \sqrt{x}$

$\lim_\limits{x \to \infty}(a+b+c) \sqrt{x} = 0 \implies (a+b+c = 0)$

score 0 · Answer 5 · answered Jun 16 '16 at 19:51

0

You can assume $x>0$ and substitute $x=1/t^2$, with $t>0$, so the limit becomes $$ \lim_{t\to0^+}\frac{a\sqrt{1+t^2}+b\sqrt{1+2t^2}+c\sqrt{1+3t^2}}{t} $$ The limit is zero if $a+b+c=0$ and infinite otherwise.

answered Jun 16 '16 at 19:51

egreg

238,574

fleablood · Answer 6 · 2016-06-16T21:26:55.297

Lessee...

first prove $\lim_{x\rightarrow \infty}\frac{\sqrt {x + j}}{\sqrt{x+k}}= 1$ which should be easy (albeit tedious) enough. [$\frac{\sqrt{x + j}}{\sqrt{x + k}} =\frac{ \sqrt{1 + j/x}}{\sqrt{1 + k/x}}$ so limit is 1].

$\lim_{x\rightarrow \infty}a\sqrt{x + 1} + b \sqrt{x+2} + c\sqrt{x+3} =$

$\lim_{x\rightarrow \infty}(a + b\frac{\sqrt{x+2}}{\sqrt{x+1}} + c\frac{\sqrt{x+3}}{\sqrt{x+1}})\sqrt{x+1}=$

$\lim_{x\rightarrow \infty}(a + \lim_{x\rightarrow \infty}b\frac{\sqrt{x+2}}{\sqrt{x+1}} + \lim_{x\rightarrow \infty}c\frac{\sqrt{x+3}}{\sqrt{x+1}})\sqrt{x+1}=$

$\lim_{x\rightarrow \infty}(a + b + c)\sqrt{x+1}=$

$(a + b + c)\lim_{x\rightarrow \infty}\sqrt{x+1}=$

$\{\sqrt{x+1}\}$ diverges. So If $a + b + c \ne 0$ then $\{(a + b + c)\sqrt{x+1}\}$ diverges.

If $a + b + c = 0$ then $\lim_{x\rightarrow \infty}(a + b + c)\sqrt{x+1}=\lim_{x \rightarrow \infty} 0 = 0.$

It's a bit frustrating because "obviously" lim root (x + 1) = lim root(x + 2) so obviously lim a(root (x + 1)) + b(root(x+2)) + c(root(x+3)) = (a+b+c)(root(x + i) = 0 iff a+ b + c = 0. But you can't say lim root(x+1) = lim root(x +2) as they diverge. — fleablood, Jun 16 '16 at 21:22

Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$

6 Answers6