<s>[INST] 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.
 [/INST] The limit is bounded if a+b+c = 0
verify:
$\lim_\limits{x \to \infty} \sqrt{x+k} - \sqrt {x} = 0$
$\lim_\limits{x \to \infty} a\sqrt{x+1} -a\sqrt{x}+ b\sqrt{x+2} -b\sqrt{x} + c\sqrt{x+3} -c\sqrt{x}= 0$
$\lim_\limits{x \to \infty} (a\sqrt{x+1} + b\sqrt{x+2} + c\sqrt{x+3}) = \lim_\limits{x \to \infty}(a+b+c) \sqrt{x}$
$\lim_\limits{x \to \infty}(a+b+c) \sqrt{x} = 0 \implies (a+b+c = 0)$
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