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