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$ 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.
 A: 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.
A: 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.
A: 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 $a+b+c=0$ 
A: The limit is bounded if a+b+c = 0
verify:
$\lim_\limits{x \to \infty} \sqrt{x+k} - \sqrt {x} = 0$
$\lim_\limits{x \to \infty} a\sqrt{x+1} -a\sqrt{x}+ b\sqrt{x+2} -b\sqrt{x} + c\sqrt{x+3} -c\sqrt{x}= 0$
$\lim_\limits{x \to \infty} (a\sqrt{x+1} + b\sqrt{x+2} + c\sqrt{x+3}) = \lim_\limits{x \to \infty}(a+b+c) \sqrt{x}$
$\lim_\limits{x \to \infty}(a+b+c) \sqrt{x} = 0 \implies (a+b+c = 0)$
A: 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.
A: 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 $\lim_{x\rightarrow \infty}(a + b + c)\sqrt{x+1}=\lim_{x \rightarrow \infty} 0 = 0.$
