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.

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.
